TITLE: Divergence in spherical coordinates vs. cartesian coordinates
QUESTION [6 upvotes]: The divergence in spherical coordinates is derived as
$$\nabla\vec A=\frac{1}{r^2}\frac{\partial(r^2A_r)}{\partial r}+\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}(A_\theta\sin\theta)+\frac{1}{r\sin\theta}\frac{\partial A_\varphi}{\partial\varphi}$$
However, this seems to fail at a simple example. Let's take the identity $\vec F(\vec r):=\vec r$ for example. The divergence of the vector field $\vec F(\vec r)=\begin{pmatrix}x\\y\\z\end{pmatrix}$ in cartesian coordinates is
$$\nabla\vec F(\vec r)=\frac{\partial}{\partial x}F_x(\vec r)+\frac{\partial}{\partial y}F_y(\vec r)+\frac{\partial}{\partial z}F_z(\vec r)=\frac{\partial}{\partial x}x+\frac{\partial}{\partial y}y+\frac{\partial}{\partial z}z=3$$
But in spherical coordinates the vector field would be fined as $\vec F(\vec r)=\begin{pmatrix}r\\\theta\\\varphi\end{pmatrix}$, so the divergence would be:
$$\nabla\vec F(\vec r)=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\cdot r)+\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}(\theta\sin\theta)+\frac{1}{r\sin\theta}\frac{\partial}{\partial\varphi}(\varphi)=3+\frac{\sin\theta+\theta\cos\theta}{r\sin\theta}+\frac{1}{r\sin\theta}$$
Which certainly isn't 3:

(Graphic from WolframAlpha)
So what am I doing wrong here?

Edit I successfully managed to completely confuse myself. I think I misunderstood the meaning of $A_r, A_\theta, A_\varphi$ - could anybody explain it to me? In my understanding the following should be true:
$$A_r=\vec A\cdot\hat e_r\qquad A_\theta=\vec A\cdot\hat e_\theta\qquad A_\varphi=\vec A\cdot\hat e_\varphi$$
With the unit vectors for spherical coordinates being as follows:
$$\hat {e} _{r}={\begin{pmatrix}\sin \theta \cos \varphi \\\sin \theta \sin \varphi \\\cos \theta \end{pmatrix}},\qquad \hat {e} _{\theta }={\begin{pmatrix}\cos \theta \cos \varphi \\\cos \theta \sin \varphi \\-\sin \theta \end{pmatrix}},\qquad \hat {e} _{\varphi }={\begin{pmatrix}-\sin \varphi \\\cos \varphi \\0\end{pmatrix}}$$
However, this doesn't make much more sense, as - if I understood it correctly, $\varphi$ and $\theta$ would be $45^\circ$ (and $r=1$) which results in weird basis vectors and doesn't seem to help me.

REPLY [2 votes]: It is sadly common when dealing with different coordinate systems, such as in this case, not to distinguish the function under consideration from its pullback under a coordinate transformation.
Pullback.
Let me introduce some terminology. Suppose we have sets $A$, $B$, and $C$ along with maps
$$A \xrightarrow{\Phi} B \xrightarrow{u} C.$$
We may then form the composition $u \circ \Phi$ of $u$ with $\Phi$, which is the mapping of $A$ into $C$,
$$A \xrightarrow{u \circ \Phi} C,$$
defined by letting
$$(u \circ \Phi)(a) = u(\Phi(a)) \quad\text{for all}\quad a \in A.$$
We call $u \circ \Phi$ the pullback of $u$ under $\Phi$, sometimes denoted $\Phi^\ast u$. Below we will also denote this map simply by $\tilde u$, thus
$$\tilde u = \Phi^\ast u = u \circ \Phi$$
will all denote the same map in the below considerations. (The trouble arises when people do not distinguish between $u$ and $\tilde u$.)
Spherical Coordinate Example.
Consider now the subset $A$ of $\mathbf{R}^3$ given by
$$A = [0, \infty) \times [0, \pi] \times [0, 2\pi] = \bigl\{ (r, \theta, \phi) : r \ge 0, 0 \le \theta \le \pi, 0 \le \phi \le 2\pi \bigr\}$$
along with the mapping $\Phi$ of $A$ into $\mathbf{R}^3$ given by
$$\Phi(r, \theta, \phi) = \begin{bmatrix} r\sin\theta\cos\phi \\ r\sin\theta\sin\phi \\ r\cos\theta \end{bmatrix}.$$
If for instance $u$ is a mapping of $\mathbf{R}^3$ into $\mathbf{R}$, we call the pullback $\tilde u = u \circ \Phi$ of $u$ under $\Phi$ the representation of $u$ in spherical coordinates.
As an example, suppose $u$ is defined by letting
$$u(x, y, z) = (x^2 + y^2 + z^2)^{\frac 12} \quad\text{for all}\quad (x, y, z) \in \mathbf{R}^3,$$
and let us compute the representation of $u$ in spherical coordinates. For a point $(r, \theta, \phi) \in A$, one has
$$\begin{align*}
\tilde u(r, \theta, \phi)
&= u(\Phi(r, \theta, \phi))
= u\bigl( r\sin\theta\cos\phi, r\sin\theta\sin\phi, r\cos\theta \bigr) \\
&= \bigl( (r\sin\theta\cos\phi)^2 + (r\sin\theta\sin\phi)^2 + (r\cos\theta)^2 \bigr)^{\frac 12} \\
&= r.
\end{align*}$$
Thus, $u$ is the mapping on $\mathbf{R}^3$ sending the point $(x,y,z)$ to $(x^2 + y^2 + z^2)^{1/2}$, whereas $\tilde u$ is the mapping on $A$ sending the point $(r, \theta, \phi)$ to $r$, i.e., projection on the first factor.
Observe that $u$ looks a lot simpler in spherical coordinates! It is therefore of interest to see if we can compute such quantities as the gradient $\nabla u$ of $u$ in terms of the gradient $\nabla\tilde u$ of $\tilde u$ and the mapping $\Phi$.
Vector Fields in Spherical Coordinates.
The representation of a vector field in spherical coordinates is worked out in the same way. Consider your example $F$ mapping a point $(x,y,z)$ in $\mathbf{R}^3$ to itself:
$$F(x,y,z) = \begin{bmatrix} x \\ y \\ z \end{bmatrix}.$$
Its pullback under $\Phi$ at a point $(r, \theta, \phi) \in A$ is given by
$$\begin{align*}
\tilde F(r, \theta, \phi)
&= F(\Phi(r, \theta, \phi)) \\
&= F\bigl( r\sin\theta\cos\phi, r\sin\theta\sin\phi, r\cos\theta \bigr) \\
&= \begin{bmatrix} r\sin\theta\cos\phi \\ r\sin\theta\sin\phi \\ r\cos\theta \end{bmatrix}.
\end{align*}$$
Note carefully that although $F$ is the identity mapping on $\mathbf{R}^3$, the representation $\tilde F$ of $F$ in spherical coordinates is not the identity mapping on $A$! However, an interesting question is whether one can recover $\operatorname{div} F$ from $\tilde F$ and $\Phi$, which is what the formula you provided is meant to do.
As for the notation «$F_r$, $F_\theta$, $F_\phi$», one means the projection of $\tilde F$ along the directions of the unit vectors $e_r$, $e_\theta$, and $e_\phi$, respectively. (It would be better to denote these by $\tilde F_r$, $\tilde F_\theta$, $\tilde F_\phi$, but we will not do so below.) Thus at a point $(r, \theta, \phi)$, one means for instance
$$F_r(r, \theta, \phi) = \tilde F(r, \theta, \phi) \cdot e_r(r, \theta, \phi),$$
where we have explicitly evaluated $e_r$ at $(r, \theta, \phi)$.
It is an exercise to show that
$$F_r(r, \theta, \phi) = r, \quad F_\theta(r, \theta, \phi) = 0, \quad F_\phi(r, \theta, \phi) = 0.$$
Using this, the formula you provide for the divergence of $F$ in spherical coordinates becomes
$$\begin{align*}
(\operatorname{div} F)(r\sin\theta\cos\phi, r\sin\theta\sin\phi, r\cos\theta)
&= (\operatorname{div} F)(\Phi(r, \theta, \phi)) \\
&= \frac{1}{r^2}\frac{\partial}{\partial r} \bigl( r^2 F_r(r, \theta, \phi) \bigr) \\
&= \frac{1}{r^2} \frac{\partial}{\partial r}\bigl( r^3 \bigr) \\
&= 3,
\end{align*}$$
which is correct.


text:
\begin{document}
\maketitle

\begin{abstract}
We study Nivat's conjecture on algebraic subshifts and prove that
in some of them every low complexity configuration is periodic. This is the case
in the Ledrappier subshift (the 3-dot system) and, more generally, in all two-dimensional
algebraic subshifts over $\FF$ defined by a polynomial without line polynomial factors in
more than one direction. We also find an algebraic subshift that is defined by a product of
two line polynomials that has this property (the 4-dot system) and another one that does not.
\end{abstract}

\section{Introduction and preliminaries}

The European Association for Theoretical Computer Science, EATCS, celebrated its 25th anniversary at ICALP 1997 in Bologna. The keynote address was given by Maurice Nivat. His talk --
titled  ``Towards a History of Automata: Why Were They Introduced? What Are They Good for?'' -- suggested the following intriguing problem that has become known as Nivat's conjecture: If an infinite two-dimensional grid has been colored in such a way that, for some $n,m\in\N$, the number of distinct $n\times m$ patterns is at most $nm$, is the coloring necessarily periodic in some direction? The conjecture has attracted wide interest, but it has turned out to be difficult and remains still unsolved today.

A number of partial results and related observations have been made. Periodicity has been established if the number of $n\times m$ patterns is at most $\alpha nm$ for progressively larger and larger constants $\alpha<1$: First for $\alpha=1/144$ in~\cite{epifanio}, then for $\alpha=1/16$ in~\cite{zamboni}, and finally for the best known constant $\alpha=1/2$ in~\cite{cyrkra}.
It is also known that having at most $2n$ patterns of size $2\times n$ and having at most $3n$ patterns of size $3\times n$ imply periodicity~\cite{sander,cyrkra16}.
In~\cite{kariszabados} we introduced an algebraic approach that leads to the following asymptotic result: A non-periodic coloring of the grid can only have finitely many pairs $(n,m)$ such that the number of distinct $n\times m$ patterns is at most $nm$. Of course Nivat's conjecture claims
that there are no such pairs.

In this paper we continue with our algebraic approach and prove that Nivat's conjecture holds on colorings
that come from certain algebraically defined sets, including the Ledrappier subshift~\cite{ledrappier}. In fact we prove a stronger statement that
in these sets periodicity is implied if for any shape -- not necessarily a rectangle -- the number of patterns of that shape is at most
the size of the shape. Even more: if a coloring has any non-trivial annihilator over $\Z$ (terms explained below) then it is periodic.

To make the statement precise we
recall some terminology. For a finite alphabet $A$, colorings $c\in A^{\Z^2}$ of the two-dimensional grid by elements of $A$ are called (two-dimensional) \emph{configurations}. Typically we use notation $c_{\vec{n}}$ for the color $c(\vec{n})\in A$ of cell $\vec{n}\in\Z^2$.
Basic operations on $A^{\Z^2}$ are \emph{translations\/}:
for any $\vec{t}\in\Z^2$ the translation $\tau_{\vec{t}}:A^{\Z^2}\longrightarrow A^{\Z^2}$ by $\vec{t}$ is defined by $\tau_{\vec{t}}(c)_{\vec{n}}=c_{\vec{n}-\vec{t}}$, for all $c\in A^{\Z^2}$ and all $\vec{n}\in\Z^2$.
We call a configuration $c$ \emph{periodic\/} if $\tau_{\vec{t}}(c)=c$ for some non-zero $\vec{t}\in\Z^2$, and we call
$\vec{t}$ a \emph{vector of periodicity}. If there are two linearly independent vectors of periodicity then $c$ is \emph{two-periodic}.
In this case it is easy to see that there are horizontal and vertical vectors of periodicity $(k,0)$ and $(0,k)$ for some $k\neq 0$, and consequently a vector of periodicity in every rational direction. We call $c$ \emph{one-periodic} if it is periodic but not two-periodic.

Colorings $p\in A^D$ of a finite shape $D\subset \Z^d$ are called \emph{$D$-patterns}, or simply patterns. The set of $D$-patterns that appear in a configuration $c$ is denoted by $P(c,D)$, that is,
$$P(c,D)=\{\tau_{\vec{t}}(c)_{|D}\ |\ \vec{t}\in\Z^2\ \}.$$
We say that $c$ has \emph{low complexity\/} with respect to shape $D$ if $|P(c,D)|\leq |D|$,
and we call $c$ a \emph{low complexity configuration\/} if it has low complexity with respect to some finite $D$.
Otherwise $c$ is a \emph{high complexity configuration}.
Low complexity configurations come up naturally in practical setups, e.g. in crystallography
particles may attach to each other only in very limited ways. Moreover, there are interesting mathematical problems associated to them, including
the famous Nivat's conjecture discussed above:
\begin{conjecture}[Maurice Nivat 1997]
Let $c\in A^{\Z^2}$ be a two-dimensional configuration. If $c$ has low complexity with respect to some rectangle $D=\{1,\dots,n\}\times\{1,\dots,m\}$ then $c$ is periodic.
\end{conjecture}
\noindent
The analogous claim  in dimensions higher than two fails, as does an analogous claim in two dimensions
for many other shapes than rectangles~\cite{cassaigne}.

To study the conjecture algebraically we replace the colors by integers, or elements of some other integral domain $R$, and express the
configuration $c$ as a formal power series $c(X,Y)$ over two variables $X$ and $Y$ in which the coefficient of monomial $X^iY^j$ is $c_{i,j}$, for all $i,j\in\Z$. Note that the exponents of the variables range from $-\infty$ to $+\infty$. In the following also polynomials may have negative powers of variables so all polynomials considered are actually Laurent polynomials. Let us denote by $R[X^{\pm 1}, Y^{\pm 1}]$ and $R[[X^{\pm 1}, Y^{\pm 1}]]$ the sets of such polynomials and power series, respectively, with coefficients in domain $R$.
We call a power series $c\in R[[X^{\pm 1}, Y^{\pm 1}]]$ \emph{finitary} if its coefficients take only finitely many different values. Since we color the grid using finitely many colors,
configurations are identified with finitary power series.

Multiplying configuration $c\in R[[X^{\pm 1}, Y^{\pm 1}]]$
by a monomial corresponds to translating it, and the periodicity of the configuration by vector $\vec{t}=(n,m)$ is then
equivalent to $(X^nY^m-1)c=0$, the zero power series. More generally, we say that polynomial $f\in R[X^{\pm 1}, Y^{\pm 1}]$ \emph{annihilates} power series $c$ if the formal product $fc$ is the zero power series.
The set of polynomials that annihilates a power series is a polynomial ideal, and is denoted by
\[ \Ann_R(c) = \{ f\in R[X^{\pm 1}, Y^{\pm 1}] ~|~ fc=0 \} .\]

We observed in~\cite{kariszabados} that, in the case $R=\Z$, if a configuration
has low complexity with respect to some shape then it is annihilated by some non-zero polynomial $f\neq 0$. The proof works unchanged
for an arbitrary field $R$.

\begin{lemma}[\cite{kariszabados}]
\label{th:low_complexity}
Let $R$ be a field or $R=\Z$. Let $c\in R[[X^{\pm 1}, Y^{\pm 1}]]$ be a low complexity configuration.
Then $\Ann_R(c)$ contains a non-zero polynomial.
\end{lemma}

One of the main results of~\cite{kariszabados} states that, in the case $R=\Z$, if a
configuration $c$ is annihilated by a non-zero polynomial then it has annihilators of particularly nice form:
\begin{theorem}[\cite{kariszabados}]
  \label{th:decompo}
  Let $c\in\Z[[X^{\pm 1}, Y^{\pm 1}]]$ be a configuration (a finitary power series) annihilated by some non-zero polynomial.\\
  Then there exists non-zero $(i_1, j_1), \hdots, (i_m, j_m)\in\Z^2$ such that
  \[ (X^{i_1}Y^{j_1} - 1) \cdots (X^{i_m}Y^{j_m} - 1) \in \Ann_{\Z}(c) .\]
\end{theorem}
\noindent
Unlike Lemma~\ref{th:low_complexity} this theorem does not generalize to all cases when $R$ is a field.

For a polynomial $f = \sum a_{i,j} X^iY^j$, we call $\text{supp}(f) = \{ (i,j) ~|~ a_{i,j}\neq 0 \}$ its
\emph{support}. A \emph{line polynomial} is a polynomial with its terms aligned all on the same line: $f$ is a line polynomial in direction $\vec{u}\in\Z^2\setminus\{\vec{0}\}$ if and only if $\text{supp}(f)$ contains at least two elements and for some $\vec{n}\in\Z^2$ we have $\text{supp}(f)\subseteq \{\vec{n}+r\vec{u}\ |\ r\in\Q\}$.
Note that the annihilator provided by Theorem~\ref{th:decompo} is a product of line polynomials. A central feature of line polynomials is that any
configuration that is annihilated by a line polynomial is periodic in the direction of the line polynomial~\cite{kariszabados}.

If $R$ is a finite field and if $f\in R[X^{\pm 1}, Y^{\pm 1}]$ is a non-zero polynomial then we define the set
$$
X_f=\{ c\in R[[X^{\pm 1}, Y^{\pm 1}]]\ |\ fc=0\}
$$
of
all configurations that $f$ annihilates
and call it the \emph{algebraic subshift} defined by $f$. It is a subshift of finite type (SFT, see~\cite{lindmarcus} for the symbolic dynamics terminology). We have that $c\in X_f\Longleftrightarrow f\in\Ann_R(c)\Longleftrightarrow fc=0$.

By Lemma~\ref{th:low_complexity} all low complexity configurations belong to algebraic subshifts when the symbols are renamed as elements of
a finite field. It is thus enough to consider Nivat's conjecture on elements of algebraic subshifts: if the conjecture is false then there is a counter example configuration
that belongs to an algebraic subshift. This is the approach taken in this work. We consider various algebraic subshifts and see whether low complexity implies periodicity
in these subshifts. When this is the case we say that the subshift has the \emph{generalized Nivat property}.
We prove that an algebraic subshift $X_f$ has the generalized Nivat property if the defining polynomial $f$ does not have line polynomial factors in two different directions.

The paper is organized as follows. In Section~\ref{sec:ledrappier} we consider as a particular example the well known Ledrappier subshift,
also known as the 3-dot system, and prove that any low complexity configuration in the Ledrappier subshift is periodic. In Section~\ref{sec:general}
we generalize this result to all algebraic subshifts where the defining polynomial does not have line polynomial factors in more than one direction. This includes, for example, the
space-time diagrams of all additive cellular automata over finite fields. In Section~\ref{sec:fourdot} we consider the 4-dot system defined by polynomial $(1+X)(1+Y)$
over $\F$ and prove that it also has the generalized Nivat property although the defining polynomial has horizontal and vertical line polynomial factors,
while the analogous system defined by $(1+X^2)(1+Y^2)$ does not have the property.

\section{The Ledrappier Subshift}
\label{sec:ledrappier}

The Ledrappier subshift, or the 3-dot system,  consists of the configurations over the binary alphabet $\F$
that are annihilated by the polynomial $1+X+Y \in \F[X^{\pm 1}, Y^{\pm 1}]$.

\begin{definition}[\cite{ledrappier}]
  The \emph{Ledrappier subshift} $L$ is the algebraic subshift $X_{f_L}$ over $\F$ defined by the annihilator $f_L = 1+X+Y$.
\end{definition}

In this work we relate the generalized Nivat property to the number of line polynomial factors of the defining annihilator.
It turns out that $f_L$ has none.

A visual way of seeing if a polynomial has a line polynomial factor is to look at the shape of the convex hull of its support: the convex hull has
parallel sides in the directions of its line polynomial factors.
To be more precise, we need some definitions of discrete geometry.
The closed half plane in a direction $\vec{v}\in\Z^2\setminus\{\vec{0}\}$ is the set
$\overline{H}_{\vec{v}}=\{\vec{x}\in\Z^2 |\ \vec{x}\cdot\vec{v}\geq 0\}$, and the open half plane
$H_{\vec{v}}$ is defined analogously to contain those $\vec{x}\in\Z^2$ that satisfy $\vec{x}\cdot\vec{v} > 0$. The boundary of the half plane
is $\overline{H}_{\vec{v}}\setminus H_{\vec{v}}$.
We say that a finite set $D\subseteq\Z^2$ has an \emph{outer edge} perpendicular to $\vec{v}\in\Z^2\setminus\{\vec{0}\}$ if there is $\vec{x}\in D$ such that $D\subseteq \vec{x}+\overline{H}_{\vec{v}}$ and there are at least two elements
of $D$ on the boundary $\vec{x}+(\overline{H}_{\vec{v}}\setminus H_{\vec{v}})$. See \cref{fig:halfplane} for an illustration.

\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.4]{img/halfplane.png}
\end{center}
  \caption{The set of black cells has an outer edge perpendicular to vector $\vec{v}$.}
  \label{fig:halfplane}
\end{figure}

\begin{lemma}
\label{lem:halfplane}
Let $g,h$ be non-zero polynomials such that $\text{supp}(g)$ has an outer edge perpendicular to $\vec{v}$. Then also
$\text{supp}(gh)$ has an outer edge perpendicular to $\vec{v}$.
\end{lemma}

\begin{proof}
Let vector $\vec{u}\in\Z^2\setminus\{\vec{0}\}$ be perpendicular to $\vec{v}$. By the hypotheses of the lemma
there are $\vec{x}\in \Z^2$ and a line polynomial $\alpha$ in the direction $\vec{u}$ such that
$\text{supp}(g)\subseteq \vec{x}+\overline{H}_{\vec{v}}$ and
$\text{supp}(g-\alpha)\subseteq \vec{x}+H_{\vec{v}}$. Here $\alpha$ contains the terms of $g$ along the boundary of the half plane $\vec{x}+H_{\vec{v}}$.
Analogously, for any non-zero polynomial $h$ there exists $\vec{y}\in \Z^2$ and polynomial $\beta\neq 0$ that is either
a monomial or a line polynomial in the direction $\vec{u}$ such that
$\text{supp}(h)\subseteq \vec{y}+\overline{H}_{\vec{v}}$ and
$\text{supp}(h-\beta)\subseteq \vec{y}+H_{\vec{v}}$. But then
$\text{supp}(gh)\subseteq \vec{x}+\vec{y}+\overline{H}_{\vec{v}}$
and $\text{supp}(gh-\alpha\beta)\subseteq \vec{x}+\vec{y}+H_{\vec{v}}$.
Because $\alpha\beta$ is a line polynomial in the direction $\vec{u}$,
this proves that the support of
$gh$ has an outer edge perpendicular to $\vec{v}$.
\end{proof}

\begin{corollary}
If $f\neq 0$ has a line polynomial factor in the direction $\vec{u}$
then $\text{supp}(f)$ has outer edges perpendicular to $\vec{v}$ and $-\vec{v}$, where $\vec{v}$ is a
vector perpendicular to $\vec{u}$.
\end{corollary}

\begin{proof}
A line polynomial $g$ in the direction $\vec{u}$ has outer edges perpendicular to $\vec{v}$ and $-\vec{v}$. The claim then follows directly from Lemma~\ref{lem:halfplane}.
\end{proof}

Because the convex hull of $\text{supp}(f_L)$ is a triangle, it does not have parallel outer edges and therefore
no line polynomial factors.

\begin{corollary}
  \label{th:ledrappier_no_line_poly}
  Polynomial $f_L = 1+X+Y$ has no line polynomial factors.
\end{corollary}

\noindent
{\bf Remark.} In fact, it is very easy to see that $f_L$ is an irreducible polynomial, which directly implies Corollary~\ref{th:ledrappier_no_line_poly}.
\medskip

\noindent
Now we can prove that the Ledrappier subshift has the generalized Nivat property.

\begin{theorem}
  \label{th:ledrappier}
  Any low complexity $c\in X_{f_L}$ is periodic.
\end{theorem}

\begin{proof}
In this proof we are going to need annihilators of $c$ over $\F$ and over $\Z$. We interpret $c$ as a configuration over $\Z$ using the renaming $\F\longrightarrow \Z$ that maps $0_\F\mapsto 0_\Z$ and  $1_\F\mapsto 1_\Z$. We write
$\Ann_\Z(c)$ for the annihilator ideal of the configuration over $\Z$ obtained after this renaming.
We are going to prove the following stronger statement: if $\Ann_\Z(c)$ contains a non-zero polynomial then $c$ is  periodic. The result then follows from Lemma~\ref{th:low_complexity}.

Let $c\in L$ be non-periodic. We first prove that $\Ann_\F(c)$ is the principal ideal generated by $f_L$, that is,
for every $g\in \Ann_\F(c)$ there exists $\alpha\in\F[X^{\pm 1}, Y^{\pm 1}]$ such that $g=\alpha f_L$.
First, let $g\in \Ann_\F(c)$ be a proper polynomial which means that all exponents of variables are non-negative.
Because $X = 1+Y+f_L$ we can eliminate variable $X$: there exists a polynomial $\alpha(X,Y)$ and a polynomial $\beta(Y)$ in variable $Y$ only such that $g = \alpha(X,Y)f_L+\beta(Y)$.
Because both $f_L$ and $g$ are annihilators of $c$, $\beta(Y)\in\Ann_\F(c)$ as well.

If $\beta\neq 0$ then it is either a single monomial (in which case $c=0$) or a line polynomial annihilator of $c$. Any configuration with a line polynomial annihilator is periodic in the direction of the polynomial. We conclude that $\beta=0$ so that $g = \alpha f_L$, as claimed.

Consider then arbitrary $g\in \Ann_\F(c)$ with possibly some negative exponents.
Because $g=X^iY^jg'$ for some $i,j\in\Z$ and a proper polynomial $g'\in\Ann_\F(c)$, we conclude that also in this case  $g$ is a multiple of $f_L$, and therefore we have that
 $\Ann_\F(c)$ is the principal ideal generated by $f_L$.

Now it remains to show that $\Ann_\Z(c)$, the set of annihilators over $\Z$, is trivial. Suppose by contrary that
there is a non-zero annihilator in $\Ann_\Z(c)$. By Theorem~\ref{th:decompo}
there exists non-zero $(i_1, j_1), \hdots, (i_m, j_m)\in\Z^2$ such that
$(X^{i_1}Y^{j_1} - 1) \cdots (X^{i_m}Y^{j_m} - 1) \in \Ann_{\Z}(c)$.
By performing computations modulo two instead, we have that
$(X^{i_1}Y^{j_1} - 1) \cdots (X^{i_m}Y^{j_m} - 1) \in \Ann_{\F}(c)$.
But then this polynomial, which only has line polynomial factors, is a multiple of $f_L$.
All factors of line polynomials are line polynomials in the same direction (or monomials) so that all
irreducible factors of $f_L$ are line polynomials, contradicting Corollary~\ref{th:ledrappier_no_line_poly}.
Here we used the fact that every polynomial can be factored uniquely into its irreducible factors~\cite{cox}.

\end{proof}

\section{Annihilators with Few Line Polynomial Factors}
\label{sec:general}
One of the key element in the proof of \cref{th:ledrappier} is the fact that we can find an annihilator with one variable eliminated from it.
It turns out that this is something we can do with other annihilators than $f_L$, and to do so we will use some parts of \textit{elimination theory}, so-called
resultants. Everything needed to understand this paper will be introduced but for more, Cox's book \cite{cox} is a good introduction.

\subsection{Resultant}
Usually resultant are defined for proper polynomials, so in order to stick with usual definitions we define them likewise. We will still use them for talking about Laurent polynomials, which is not a problem since from a Laurent polynomial annihilator we can always obtain a proper one by multiplying it by a suitable monomial.

Let $\K$ be a field, e.g. $\K=\FF$ for a prime $p$, and
let $K = \K[Y]$ be the ring of polynomials with one variable $Y$. Then $K[X] =\K[X,Y]$ is the ring of polynomials with two variables over $\K$.
Resultant can be defined for polynomials with more than two variables, but two is all we need here.
Resultant can be defined formally as a determinant of some matrix with coefficients in $K$, but for our purpose, all we need to know is that for any $f,g\in \K[X,Y]$, there exists a polynomial $\Res(f,g)\in \K[Y]$ in variable $Y$ only with the following two properties:

\begin{prop}
  \label{prop:det0}
  The polynomials $f$ and $g$ have a common factor in $\K[X,Y]$ if and only if $\Res(f,g) = 0$.
\end{prop}

\begin{prop}
  \label{prop:det}
  There exist $\alpha, \beta\in \K[X,Y]$ such that
  \[ \alpha f + \beta g = \Res(f,g) .\]
\end{prop}

The second property exactly states that we can find a linear combination of $f$ and $g$ that eliminates variable $X$.

\begin{lemma}
  \label{th:two_ann}
  Let $c$ be a power series over a field $R$. If $c$ is annihilated by two non-zero polynomials $f$ and $g$ with no common factors then $c$ is two-periodic.
\end{lemma}

\begin{proof}
Using \cref{prop:det}, let $\alpha, \beta\in \K[X,Y]$ be such that
\[ \alpha f + \beta g = \Res(f,g) .\]
Because $f$ and $g$ are both annihilators of $c$ so is $\Res(f,g)$.
Because $f$ and $g$ have no common factors, \cref{prop:det0} tells us that $\Res(f,g)\neq 0$.
We found an annihilator which is a line polynomial or a monomial (only with variable $Y$) so $c$ must be periodic along $Y$.

Symmetrically, there also exist $\gamma, \delta\in \K[X,Y]$ such that
\[ \gamma f + \delta g = \ResY(f,g) .\]
Again,  $\ResY(f,g)\neq 0$, so $c$ is also periodic along $X$.
\end{proof}

\noindent
{\bf Remark.}
In~\cite{karifull} we analyzed the structure of annihilator ideals of configurations over $\Z$, and an analogous result for the case of $\Z$  follow from those considerations.
We included here the simple proof above for the sake of completeness, and in order to cover also the case of arbitrary coefficient fields.

\subsection{Main Result}

Now we can easily generalize \cref{th:ledrappier}. We consider an algebraic subshift $X_f$ over a finite field $\FF$ defined by an annihilator
$f\in \FF[X^{\pm 1}, Y^{\pm 1}]$. A configuration $c\in \FF[[X^{\pm 1}, Y^{\pm 1}]]$ will also be interpreted as a configuration over $\Z$
by mapping the symbols by $a_{\FF} \mapsto a_{\Z}$ for all $a\in\{0,1,\dots ,p-1\}$. Then, as in the proof of \cref{th:ledrappier}, we can define both
$\Ann_\FF(c)$ and $\Ann_\Z(c)$, and use the fact that any $g\in\Ann_\Z(c)$ is also in $\Ann_\FF(c)$ when its coefficients are reduced modulo $p$.

\begin{theorem}
  \label{th:main_th}
  Let $c\in X_f$ for a polynomial $f\in \FF[X^{\pm 1}, Y^{\pm 1}]$, and suppose that $\Ann_\Z(c)$ contains a non-zero polynomial.
  \begin{itemize}
    \item If $f$ has no line polynomial factors then $c$ is two-periodic.
    \item If all line polynomial factors of $f$ are in the same direction then $c$ is periodic in this direction.
  \end{itemize}
\end{theorem}
\begin{proof}
By Theorem~\ref{th:decompo}
there exists non-zero $(i_1, j_1), \hdots, (i_m, j_m)\in\Z^2$ such that
$(X^{i_1}Y^{j_1} - 1) \cdots (X^{i_m}Y^{j_m} - 1) \in \Ann_{\Z}(c)$.
By performing computations modulo $p$ instead, we have that
$g(X,Y)=(X^{i_1}Y^{j_1} - 1) \cdots (X^{i_m}Y^{j_m} - 1) \in \Ann_{\FF}(c)$.

If $f$ has no line polynomial factors then $f$ and $g$ do not have any common factors. By Lemma~\ref{th:two_ann}
then $c$ is two-periodic. This proves the first claim.

Suppose then that all line polynomial factors of $f$ are in the same direction. Let $h$ be the greatest common divisor of $f$ and $g$
so that we can write $f=f'h$ and $g=g'h$ where $f'$ and  $g'$ do not have common factors. Note that $h$ is a line polynomial: as a factor of $g$
$h$ is a product of line polynomials, and as a factor of $f$ these line polynomials are all in the same direction.

Because $c'=hc$ is annihilated by both $f'$ and $g'$ it follows from Lemma~\ref{th:two_ann}
that $c'$ is two-periodic. In particular, there is a line polynomial $h'$ in the direction of $h$ that annihilates $c'$.
We have $hh'\in \Ann_{\FF}(c)$ so that $c$ is annihilated by the line polynomial $hh'$ and is therefore periodic in this direction.

\end{proof}

Using Lemma~\ref{th:low_complexity} we now immediately get that algebraic subshifts defined by an annihilator with at most one line polynomial factor
have the generalized Nivat property.

\begin{corollary}
  \label{coro:alg_sft}
  Let $c\in X_f$ for a polynomial $f\in \FF[X^{\pm 1}, Y^{\pm 1}]$ whose line polynomial factors are all in the same direction.
  If $c$ has low complexity then it is periodic.
\end{corollary}

\noindent
{\bf Remark.}
Elements of the Ledrappier subshift are exactly the space-time diagrams of the one-dimensional XOR cellular automaton. More generally,
the algebraic subshift $X_f$ defined by a polynomial of the form $f(X,Y)=Y-g(X)\in \FF[X^{\pm 1}, Y^{\pm 1}]$ consists exactly of the
space-time diagrams of the one-dimensional additive cellular automaton over $\FF$ whose local rule is given by $g(X)$. See Section~9 of~\cite{karisurvey} for a short discussion about additive cellular automata. If $g(X)$ has at least
two terms then the support of $f$ has triangular shape and therefore $f$ has no line polynomial factors. If $g(X)$ has one term then $f$ itself is
a line polynomial. In any case, Theorem~\ref{th:main_th} and Corollary~\ref{coro:alg_sft} hold for the space-time diagrams of all
one-dimensional additive cellular automata over field $\FF$.

\section{Square Annihilators}
\label{sec:fourdot}
After Corollary~\ref{coro:alg_sft}, it is natural to take a look at configurations annihilated by polynomials with line polynomial factors in more directions. It turns out that already products of two line polynomials include examples with and without the generalized Nivat property.
We first prove that the 4-dot system defined by $(1+X)(1+Y)$ over $\F$ has the generalized Nivat property and then we show that
the system defined by $(1+X^2)(1+Y^2)$ does not.

\subsection{The 4-dot system}

\begin{definition}
  The \emph{4-dot system} $S$ is the algebraic subshift $X_{f_S}$ over $\F$ defined by the annihilator $f_S =  1+X+Y+XY = (1+X)(1+Y)$.
\end{definition}

\begin{theorem}
  \label{th:square}
  Every low complexity $c\in X_{f_S}$ is periodic.
\end{theorem}

\begin{proof}
Similarly as before, we are going to prove the more general statement that if $c$ has a non-trivial annihilator $p_0$ over $\Z$ then it is periodic.

We first observe that $c=h+v$ for $h,v\in \F[[X^{\pm 1}, Y^{\pm 1}]]$  that are
$(1,0)$-periodic and $(0,1)$-periodic, respectively. Indeed, we can take $h_{i,j}=c_{0,j}$ and $v_{i,j}=c_{i,0}+c_{0,0}$, for all $i,j\in \Z$.
Because $(1+X^i)(1+Y^j)$ is a multiple of $(1+X)(1+Y)$ over $\F$, polynomial $(1+X^i)(1+Y^j)$ annihilates $c$, for all $i,j\in\Z$. This
means that $c_{i,j}=c_{0,j}+c_{i,0}+c_{0,0}=h_{i,j}+v_{i,j}$.

Using the periodicity of $h$ and $v$  we can write $h = \ones(X) s(Y)$ and $v = t(X) \ones(Y)$, with $\ones(X) = \sum_{i\in\Z} X^i$ and $s,t$ two formal series depending only one one variable.
Let us define another binary configuration $d$ by
\[ d(X,Y) = t(X)s(Y) .\]
In other words, $d$ is the configuration that has ones where both $h$ and $v$ have ones:
\[ d_{i,j} =
\begin{cases}
  1 \text{ if } h_{i,j} = v_{i,j} = 1\\
  0 \text{ otherwise}
\end{cases}
.\]
Interpreted in \Z, we have
\[ c = h+v - 2d .\]
This is the case since the two sides are identical modulo two and both sides only contain values 0 and 1.

Consider next the polynomial
\[ p = p_0 (X-1)(Y-1)\]
over $\F$. Because $p_0$ annihilates $c$, and $X-1$ and $Y-1$ annihilate $h$ and  $v$, respectively, we have that $pc=ph=pv=0$. Therefore
$pd=0$ as well, which can be written as
\[ p(X,Y) t(X) s(Y) = 0, \]
emphasizing the variable dependencies of the polynomials. We have the following two cases:
\medskip

\noindent
\emph{Case 1:} Suppose that $p(X,Y) t(X)=0$. In $p(X,Y)$ we collect together terms with the same power of variable $Y$, obtaining
$$
p(X,Y)=\sum_{j\in \Z} Y^jp_j(X)
$$
where at least some $p_j(X)$ is a non-zero polynomial. We have
$$\sum_{j\in \Z} Y^jp_j(X)t(X)=0.$$
This is an identity of formal power series so that
$p_j(X)t(X)=0$ for all $j\in\Z$. But then also $p_j(X)v=p_j(X)t(X)u(Y)=0$, so that $v$ is annihilated by a non-zero
horizontal line polynomial (or a non-zero monomial) $p_j(X)$. We conclude that
$v$ is horizontally periodic. But then also $c=h+v$ is horizontally periodic as a sum of two horizontally periodic configurations.

\medskip

\noindent
\emph{Case 2:} Suppose that $p(X,Y) t(X)\neq 0$. Now we collect in $p(X,Y) t(X)$ together variables with the same power of variable $X$, obtaining
$$
p(X,Y) t(X) = \sum_{i\in \Z} X^i q_i(Y),
$$
where at least some $q_i(Y)$ is a non-zero polynomial. Note that all $q_i(Y)$ are polynomials because powers of the variable
$Y$ only come from the polynomial $p(X,Y)$. Because
$$
0=p(X,Y)d(X,Y)=p(X,Y) t(X)s(Y) = \sum_{i\in \Z} X^i q_i(Y)s(Y),
$$
we have that $q_i(Y)s(Y)=0$ for all $i\in\Z$. Analogously to case 1 above, this implies that $h$ is vertically periodic, and therefore also $c$ is vertically periodic.
\end{proof}

\subsection{An algebraic subshift without the generalized Nivat property}

For some polynomials with two line polynomial factors, the associated subshift does not have the generalized Nivat property. This is typically the case when the annihilating polynomial allows the use of sublattices.

\begin{theorem}
  \label{th:square2}
  There exists a configuration $c$ over $\F$ annihilated by $f_T = (1+X^2)(1+Y^2)$ which is not periodic but has low complexity.
\end{theorem}

\begin{proof}
Let us take $c = h + v$, with
\begin{center}
  $ h_{i,j} =
  \begin{cases}
    1 \text{ if } j = 0 \text{ and } i \text{ even} \\
    0 \text{ otherwise}
  \end{cases}
  $
  and ~
  $ v_{i,j} =
  \begin{cases}
    1 \text{ if } i = 1 \text{ and } j \text{ even} \\
    0 \text{ otherwise}
  \end{cases}
  .$
\end{center}
Visually, $c$ is the superposition of a horizontal and a vertical line on two disjoint sublattices, see \cref{fig:sublat}.
Clearly $h$ is one-periodic with periodicity vector $(2,0)$ and $v$ is one-periodic with periodicity vector $(0,2)$. Their sum  $c$ is not periodic.

\begin{figure}[ht]
  \centering
  \scalebox{0.6}{\input{img/sublat}}
  \caption{Sublattices of $c$ and shape $D$ superimposed. The horizontal line is from $h$ and the vertical one from $v$.}
  \label{fig:sublat}
\end{figure}

The periodicity of $h$ and $v$ directly implies that $f_T$ annihilates $c$:
$h$ being $(2,0)$-periodic $(1+X^2) h = 0$ and, analogously, $v$ being $(0,2)$-periodic $(1+Y^2) v = 0$. This means that
$f_Tc=(1+X^2)(1+Y^2)h+(1+X^2)(1+Y^2)v=0$.

\begin{figure}[ht]
  \centering
  \scalebox{0.6}{\input{img/D}}
  \caption{The shape $D$ of low complexity.}
  \label{fig:D}
\end{figure}

The last thing we have to check is that $c$ has low complexity, i.e, there is a shape $D$ such that $P_c(D) \leq |D|$.
It is sufficient to take $D$ to be the scattered 3x3 square, as shown in \cref{fig:D}.
Patterns of shape $D$ in $c$ will only contain values from one of the four sublattices, depending on the parity of its position.
If $D$ is superimposed with sublattices 3 or 4, the pattern is blank.
With sublattice 1, it can only contain values from $h$, so it can have four different values: blank, and the horizontal line crossing at the top, the middle or the bottom. If it is on sublattice 2, then it has values from $v$, and here again there are four different possibilities.
Counting the blank shape only once, we obtain $P_c(D) = 1 + 3 + 3 = 7 < 9 = |D|$.

\end{proof}

\section{Conclusions}

To prove Nivat's conjecture it is enough to consider configurations of two-dimensional algebraic subshifts over $\FF$. We prove that simplest
subshifts do not contain a counter example to the conjecture, and they even have the stronger property that any configuration that has low complexity with respect to any finite shape is periodic. It remains open to characterize which algebraic subshifts have this generalized Nivat property. It remains also an interesting algorithmic problem to decide for a set of at most $|D|$ allowed patterns of shape $D$,
for some finite  $D\subseteq\Z^2$, whether there is a configuration all of whose $D$-patterns are among the allowed patterns. We do not know if
this emptyness problem of low complexity subshifts of finite type is decidable in general, but if the given patterns come from one of the algebraic subshifts studied
in this work then the emptyness problem is decidable by a standard argument because any configuration with these patterns must be periodic. We also ask
whether the generalized Nivat property may fail for other reasons than having the support of the defining polynomial in a proper sublattice of $\Z^2$.

\bibliographystyle{abbrv}
\bibliography{biblio}

\end{document}

config: arxiv


TITLE: Is $T$ necessarily a linear transformation?
QUESTION [2 upvotes]: If $T:\mathbb{R}^2 \rightarrow \mathbb{R}$ is a function such that $T(\alpha v)=\alpha T(v)$ $\forall \alpha \in \mathbb{R}$ and $v \in \mathbb{R}^2$, is T necessarily a linear transformation?
My instinct is no since surely it has to also be additive in order to be linear (by definition), but I can't seem to think of a counterexample.
Any pointers would be much appreciated!

REPLY [1 votes]: As an alternative, define
$$
T(x, 0) = (x, 0)
$$
and
$$
T(x, y) = (0,0)
$$
for $y \ne 0$. Now consider that $(1, 1) = (0, 1) + (1, 0)$ and apply $T$ to both sides assuming linearity, and arrive at a contradiction.


TITLE: Quantifier elimination, $(\mathbb{R}, <)$
QUESTION [1 upvotes]: I have a general question about quantifier elimination.
Which kinds of formulas do you have to observe?
For example let T be the theory of $(\mathbb{R}, <)$ and I want to show, that this theory has quantifier elimination.
What do I have to do in general to proof it.
Are there just specific formulas I have to observe?
Does quantifier elimination always have the same steps, so you always observe the same formulas and show that they are equivalent to a quantifier free formula?
Thanks in advance.
Edit: Spelling improved.

REPLY [2 votes]: In proving quantifier elimination, it suffices to treat formulas consisting of a single existential quantifier applied to a conjunction of atomic and negated atomic formulas.  But that is, as far as I know, the only general reduction available.  After that, you need to consider the particular theory and do some real work.


TITLE: Poincaré polynomial and blow-up
QUESTION [2 upvotes]: Consider $X,Z$ smooth projectives varieties and $Y = Bl_Z(X)$ the blow-up of $X$ with center $Z$. Finally let $E$ be the exceptional divisor of the blow-up. Is is true that $p(Y) - p(E) = p(X) - p(Z)$ where $p$ is the Poincaré polynomial ?
Thanks in advance !

REPLY [2 votes]: Yes, both sides compute the Poicare polynomial of cohomology with compact supports of $Y \setminus E = X \setminus Z$.


TITLE: Proportionality and Linearity
QUESTION [3 upvotes]: I have a very basic question that a friend and I are disagreeing on.
Does (direct) proportionality between two terms a and b imply linearity between the two terms?

REPLY [3 votes]: Yes, $a$ is directly proportional to $b$ means that $a = k b$ for some $k$.  This is a linear relationship.  Imagine something like force and acceleration.  If I double the force on an object, I double the acceleration.  If I have two different forces acting on an object, the effect is the same as one bigger force corresponding to the addition.


TITLE: Sequence of Random Variables
QUESTION [0 upvotes]: Given
$$\displaystyle \mathbb P(X_n=1)=\frac{1}{n^{\alpha}},~~~\text{and}~~~\displaystyle \mathbb P(X_n=0)=1-\frac{1}{n^{\alpha}}$$ where in $\alpha>0$,  $n\geq1$.
How can I prove that $$X_n\rightarrow 0~~\text{in}~~~ \mathcal L^{1}$$
Also, how do I show that $$\displaystyle \limsup_{n\rightarrow +\infty}X_n=
\begin{cases}
1, & \text{if }\alpha \leq 1 \\
0, & \text{if }\alpha >1
\end{cases}$$
Thanks for any help.

Attempt: The sequence converges in probability to $0$ if and only if $\displaystyle \lim_{n\rightarrow \infty}\left(1-\frac{1}{n^{\alpha}}\right)=0$.
By Borel-Cantelli Lemma the sequence converges almost surely to 0 if and only if
$\displaystyle \sum_{n=1}^{\infty}\left(1-\frac{1}{n^{\alpha}}\right)<\infty$

REPLY [1 votes]: For the first question, it's easy you need to show that
$$ \mathbb E \vert X_n - 0 \vert \longrightarrow 0.   $$
After, for $\alpha > 1$, we use Borel-Cantelli:
$$ \sum_{n=1}^\infty \mathbb P(X_n = 1) = \sum_{n=1}^\infty  \frac 1 {n^{\alpha}} < \infty   $$
to deduce that $\mathbb P(\lim \sup \{ X_n = 1 \}) = 0$ and so $ \lim \sup X_n =0$ a.s.
For the case $\alpha \leq 1$, we can also use Borel-Cantelli but we need the $X_n$'s to be independent and I don't think it is true otherwise.


TITLE: What is the zero of a group ring?
QUESTION [0 upvotes]: Let G be a finite group and R a commutative unital ring. Let RG be the corresponding group ring. My question is, does it follow that 0g = 0$1_G$ for all g in G (where $1_G$ means the identity element of the group)?
On wikipedia, it reads 'The additive identity element is of course zero'. I take that to mean that the additive identity element is 0$1_G$, which certainly has the zero property of x + 0 = x for all x. I believe the elements  0g also have this property, that is why I'm starting to think 0g = 0$1_G$ for all g in G. However, I have not been able to prove this.

REPLY [0 votes]: Any module $M$ has an additive identity that is equal to $0x$ for any $x\in M$. This is because
$$0x=(0+0)x=0x+0x$$
so
$$0x=0x-0x=0$$
It thus makes the most sense to treat $0$ as a "module thing." It behaves the same way in any module, and there's no virtue in writing it as $01_G$ or as $0x$ for any other element $x$.


TITLE: Why is the central maximum white when white light passes through a diffraction grating?
QUESTION [0 upvotes]: I can't see how path difference and central maximum (where path difference is 0) work when there are more than 2 wave sources.
When a beam of white light passing through a diffraction grating, it is obvious that the central maximum is a white fringe, while other bright fringes are arranged in spectra (due to the composition of white light).
However, I can't understand how the central maximum can be like this. When there are more than 2 circular wave sources, there will be no points where the path difference is 0 for all the waves. Therefore, there should be no path where the antinodal lines from all coloured lights overlap to produce a white fringe.
Can someone please point out what is wrong with my reasoning above?

REPLY [1 votes]: We are using a far field approximation.  Each groove in the diffraction grating acts a source of circular waves and it's true that the distances from each groove to the central maximum are different.  But once you get far enough away from the grating these differences in path length become much less than the wavelength of light.  So while light from each groove isn't travelling exactly the same distance to the central maxima and thus won't arrive exactly in phase it's close enough to produce constructive interference.


text:
\begin{document}
\maketitle
\begin{abstract}
The notion of position operator for massless spinning particles is discussed in some detail. The noncommutativity of coordinates is related to the gauge symmetry following from the freedom in definition of standard state in Wigner's construction of induced representations of Poincare group. The counterpart of Newton-Wigner operator is discussed. It is explained why the Newton-Wigner construction works only for helicity $ \mid\lambda\mid\leq1/2$.
\end{abstract}
\section{Introduction}
Particle localizability is a well defined notion in nonrelativistic quantum mechanics.  However, it lacks the precise meaning in relativistic (quantum) regime. The crucial property of relativistic quantum theory is the absence of nontrivial  interactions admitting particle number conservation. This has to be carefully taken into account when particle localization procedures are analyzed\cite{b1}. The main restriction comes from the fact we are dealing with a single particle while the one-particle relativistic quantum theory is only an approximation valid, for example, for weak and slowly varying external fields. However, once the validity of such approximation is assumed one can construct a number of candidates for particle position operator.

\par
The most reasonable proposal has been given by Newton and Wigner in their seminal work \cite{b2}, anticipated to some extent by Price \cite{b3}. The localizability problem has been further discussed in a number of papers \cite{b4}$\div$\cite{b10}, to mention only a few.
\par
One can argue that any reasonable position operator should have the following properties:
\begin{itemize}
\item[(i)] one would like to retain the interpetation of momentum operator $\hat{\vec{P}}$ as generator of translations in space; as a result the following commutation rule is assumed \\
$$ [\hat{X_i}, \hat{P^k}]=i\delta^k_i   ;$$ \\
\item[(ii)]the coordinate operator $ \hat{X_i} $ should transform as a threevector under rotations; \\
\item[(iii)] the existence of coordinate representation is assumed which implies the commutation rules
$$ [\hat{X_i}, \hat{X_k}]=0$$
\end{itemize}
It was realized some time ago that one can hardly expect all the above conditions to be fulfilled in
the case of spinning massless particles. In fact, already Newton and Wigner noticed that their construction fails to work in the limit of vanishing mass, except for helicities $ \lambda=0 $ and $ \mid\lambda\mid=1/2 $ ( in the latter case both helicities must be included). A simple explanation of this fact is essentially contained in a very nice paper by Skagerstam \cite{b11}. Basically, the reason that Newton-Wigner algorithm does not work for $ \mid\lambda\mid>1/2 $ is that, given the position operator obeying (i) $ \div $ (iii), one can construct $ SU(2) $ spin variables which is, in turn, only possible if all helicities $-\mid \lambda \mid, -\mid \lambda \mid+1,...,\mid \lambda \mid-1, \mid \lambda \mid $ are included. This Mukunda-Skagerstam argument is explained in some detail in Sec. II. In Sec. III we sketch the construction of representations of Poincare group for massless particles from the point of view of gauge symmetry following from the freedom in the choice of standard state used in Wigner's construction of induced representations. Such an approach  provides a convenient framework for discussing the problem of noncommutativity of coordinates. Sec. IV is devoted to the discussion of Newton-Wigner coordinate for parity conserving helicity $1/2$ massless theory. We show that in this special case the coordinate operator can be identified with covariant ( with respect to the gauge symmetry mentioned above ) derivative which describes trivial connection. This is a consequence of dealing with reducible representation of Poincare group encompassing both helicities. The nontrivial connections corresponding to both helicities are obtained from the trivial one by abelian projection. In Sec. V photons as an example of  $ \mid \lambda \mid=1 $ are considered. Again the covariant derivatives for both helicities are obtained by abelian projection from $ SU(2) $ connection  which, in contrast to $ \mid \lambda \mid=1/2 $ case, is also nontrivial; its nontriviality is related to the transverality  condition, i.e. lacking of $ \lambda=0 $ helicity component ( in accordance with Mukunda-Skagerstam argument ). The nontriviality of the connection results in breaking (iii) for photon coordinate operator. Sec. VI is devoted to some conclusions. In Appendix we discuss the source of nonminimal (from the point of view of gauge symmetry) terms in Poincare generators. We sketch also the construction of position operators which obey (i) and (iii) but transform nonlinearly under rotations.

\section{Localizability of massless particles}

Consider a free particle described by an irreducible representation of Poincare group. The group generators $P_\mu$ ( fourmomentum ), $\vec{J}$ (angular momentum ) and $\vec{K}$ (boosts ) are represented by observables acting in the space of states and obeying the algebra

\begin{equation}
\label{e1}
\begin{split}
&[J_i,J_k]=i\varepsilon_{ikl}J_l,\quad \quad\;\;[J_i,P_k]=i\varepsilon_{ikl}P_l,\\
&[J_i,K_k]=i\varepsilon_{ikl}K_l , \quad \quad[K_i,P_k]=i\delta_{ik}P_0, \\
&[K_i,K_k]=-i\varepsilon_{ikl}J_l ,  \quad \;[K_i,P_0]=iP_i
\end{split}
\end{equation}
with the remaining commutators vanishing.\\
Assume one can construct the observable $\vec{X}$ representing the particle position. The basic conditions $(i)\div(iii)$ from Introduction imply the following commutation rules:

\begin{equation}
\label{e2}
\begin{split}
&[X_i,P^k]=i\delta_i^k\\
&[J_i,X_k]=i\varepsilon_{ikl}X_l\\
&[X_i,X_k]=0
\end{split}
\end{equation}
Let us define the orbital angular momentum operator

\begin{equation}
\label{e3}
L_i\equiv\varepsilon_{ijk}X_jP^k
\end{equation}
together with the spin operator
\begin{equation}
\label{e4}
S_i \equiv J_i-L_i
\end{equation}
Eqs. $(1)\div(4)$ imply

\begin{equation}
\label{e5}
\begin{split}
&[S_i,S_k]=i\varepsilon_{ikl}S_l\\
&[S_i,P_k]=0
\end{split}
\end{equation}

Therefore, our space of states carries some representation of $sU(2)$ algebra spanned by the spin operators $S_i$.
\par
Consider now the case of massless particles. The relevant representations of Poincare group are uniquely determined by the choice of (half)integer helicity $\lambda$ defined by

\begin{equation}
\label{e6}
\lambda \equiv \frac{\vec{P}}{\mid \vec{P} \mid}\cdot\vec{J}=\vec{S}\cdot\frac{\vec{P}}{\mid \vec{P} \mid}
\end{equation}

Due to the second eq.(\ref{e5}) one can diagonalize $\vec{S}\cdot\frac{\vec{P}}{\mid \vec{P} \mid}$ within any subspace carrying irreducible representation of spin algebra. Therefore, the spectrum of $\vec{S}\cdot\vec{P}$ contains, together with $\lambda$, all values $-|\lambda|, -|\lambda|+1,...,|\lambda|-1,|\lambda|$.\\
This is, however, in contradiction with the operator relation (\ref{e6}). This nice argument has been given by Skagerstam \cite{b11} who ascribed it to N. Mukunda.
\par
It is now quite easy to understand why Newton-Wigner scheme works in the massless case only for $|\lambda|=0,1/2$. The spinless case is trivial. For $|\lambda|=1/2$, the values $\lambda=\pm1/2$ cover the whole spectrum of $\vec{S}\cdot\frac{\vec{P}}{\mid \vec{P} \mid}$. Therefore, there is no contradiction provided we consider the reducible representation including both helicities which is the case for Dirac equation considered in \cite{b2}. For $|\lambda|>1/2$ one would be forced to consider the irreducible representations of Poincare group describing particles of different absolute values of helicities.

\section{Massless representations from gauge principle}
\label{s3}
The construction of irreducible representations of Poincare group was described in the seminal paper of Wigner \cite{b12} and repeated in numerous articles and books. It is technically slightly involved; in the case of massless particles additional difficulty comes from the fact that the relevant orbit in momentum space is topologically nontrivial.
\par
One can look at the problem of massless representations from slightly different point of view. We start with the simplest representation carried by the complex valued functions $f(\vec{p})$ equipped with a scalar product
\begin{equation}
\label{e7}
(f,g)\equiv\int\frac{d^3\vec{p}}{p^0}\overline{f(\vec{p})}g(\vec{p}), \quad p^0=|\vec{p}|
\end{equation}
The Poincare group acts by $f(\vec{p})\rightarrow f(\Lambda^{-1}\vec{p})e^{ipa}$ which yields irreducible representation corresponding to vanishing mass and helicity. The group generators read

\begin{equation}
\label{e8}
\begin{split}
&P_\mu=p_\mu\\
&J_i=i\varepsilon_{ijk}\frac{\partial}{\partial p^j}p^k \\
&K_i=\frac{1}{2}\left(p^0(i\frac{\partial}{\partial p^i}- \frac{i}{2}\frac{p^i}{(p^0)^2}) +
(i\frac{\partial}{\partial p^i}- \frac{i}{2}\frac{p^i}{(p^0)^2})p^0\right)=ip^0\frac{\partial}{\partial p^i}
\end{split}
\end{equation}

Consider now the case of nonvanishing helicity, $\lambda\neq0$. One selects some standard fourvector
$\underline{k}$ (typically, $\underline{k}=(k,0,0,k)$) and standard boost $L(p)\in SO(3,1)$ obeying $L^\mu_{\;\;\nu}(p)k^\nu=p^\mu$.The basic vectors are defined by

\begin{equation}
\label{e9}
|p \rangle = U(L(p))|\underline k \rangle
\end{equation}
where $|\underline k \rangle $  spans one dimensional holomorphic representation of stability subgroup of $\underline k$ ( which is isomorphic to $E(2)$). The choice of the standard boost is not unique. One can make replacement $L(p)\rightarrow L(p)L_0(p)$, where $L_0(p)$ is any element of stability subgroup of $\underline k , L_0(p)\underline k =\underline k $. It is crucial to note that $L_0$ may depend on final momentum $\vec{p}$. The corresponding modified basic vector reads

\begin{equation}
\label{e10}
\begin{split}
|p \rangle' = &U(L(p)L_0(p))|\underline k \rangle=U(L(p))U(L_0(p))|\underline k \rangle = \\
&e^{i \lambda\varphi(p)}U(L(p))|\underline k \rangle=e^{i \lambda\varphi(p)}|\underline p \rangle
\end{split}
\end{equation}

which induces the gauge transformations for amplitudes $f(\vec{p}) \equiv \left\langle \vec{p} |f\right\rangle $:
\begin{equation}
\label{e11}
f'(\vec {p}) = e^{-i \lambda\varphi(\vec {p})}f(\vec {p})
\end{equation}
This gauge transformation corresponds merely to the change of conventions so it yields unitarily equivalent representation; in particular, the structure of Lie algebra is invariant under the gauge transformations(\ref{e11}). This implies that one has to replace the derivatives $i\frac{\partial}{\partial p^k} \equiv i\partial_k$ entering the generators (\ref{e8}) by their covariant counterparts \cite{b11}
\begin{equation}
\label{e12}
i{\mD}_k\equiv i \partial_k-{\mA}_k ,
\end{equation}

where the gauge field ${\mA}_k$ transforms as follows

\begin{equation}
\label{e13}
{\mA}'_k(p) = {\mA}_k(p) +\lambda \partial_k \varphi(p)
\end{equation}
It remains to determine the gauge class of ${\mA}_k(p)$. This problem can be solved by taking a closer look at Wigner construction (for example, by comparing the action of stability subgroups corresponding to opposite standard (three)momenta). As a result one concludes that the massless particles of helicity $\lambda$ are describes by line bundle over the sphere $S^2 (|\vec{p}|=$ const)
characterized by the integer $2\lambda$. Noting additionally that the Lie algebra commutation rules should be explicitly covariant under rotations one finds that ${\mA}_k$ corresponds to the field strength of monopole with "magnetic" charge $\lambda$  \cite{b11},\cite{b22}, \cite{b13},\cite{b23} . However, this is not the end of the story.  The "minimal" coupling $i\partial_k \rightarrow i{\mD}_k$ yields generators still obeying $\vec{P}\cdot\vec{J}=0$. Therefore, the definition of $\vec{J}$ must be supplied by nonminimal term producing the correct relation (\ref{e6}). Finally one obtains

\begin{equation}
\label{e14}
\begin{split}
&J_i=i\varepsilon_{ijk}D_jp^k + \lambda \frac{p^i}{|\vec{p}|}\\
&K_i=\frac{1}{2}\left(p^0(iD_i- \frac{i}{2}\frac{p^i}{(p^0)^2}) +
(iD_i- \frac{i}{2}\frac{p^i}{(p^0)^2})p^0\right)=ip^0 D_i
\end{split}
\end{equation}
One easily checks that the choice (\ref{e14})  yields correct commutation rules.
\par
Coming back to the definition of coordinate operator $X_i$  note that the first equation  (\ref{e2}) implies that $X_i$ must contain $i\partial_i$. Therefore, demanding that $X_i$ does not depend on the choice of conventions ( the gauge choice ) we conclude that $i\partial_i$ enters coordinates through covariant derivative $iD_i$ \cite{b11}.The natural choice is simply $X_i= i{\mD}_k$ Then the coordinates necessarily do not commute. Taking into account that the potential $\vec{\mA}(\vec{p})$ describes a monopole of charge $\lambda$ one immediately finds

\begin{equation}
\label{e15}
[X_i, X_j] = -i\lambda \varepsilon_{ijk} \frac{p^k}{(p^0)^3}
\end{equation}
Therefore, (iii) is not fulfilled. Let us note that the non-commutativity of the position operator can be derived from the classical description of spinning particles \cite{b24}, \cite{b25}.
\par
 On the other hand, one finds that $X_i$'s transform as the components of threevector,
\begin{equation}
\label{e16}
[J_i, X_j] = i\varepsilon_{ijk}X_k
\end{equation}
so (ii) holds. This is slightly surprising because one cannot find explicitly rotationally invariant potentials describing the field of monopole. It is interesting to consider this problem in some detail. The description of symmetries in gauge theories demands some care \cite{b14}. In fact, a given transformation is a symmetry if its "unwanted" effect can be cancelled by a suitable compensating gauge transformation. In our case the change of vector potential under rotations is cancelled by $U(1)$ gauge transformation (see Appendix for some details). As a result the initial generators must be supplied  by additional pieces which generate the relevant compensating gauge transformations ( this is the origin of well known "spin from isospin" phenomenon \cite{b15}). In the case under consideration
the second term on the right hand side of the first eq. (\ref{e14}) may be viewed  as a generator of compensating gauge transformation.
\par
Obviously, the choice $X_i=i{\mD}_i$ is not unique. If one insists on commuting coordinates, $X_i$ must be modified; then (ii) is no longer fulfilled. An example of such modified coordinate is considered in Ref. \cite{b16}. The coordinate observables are there obtained by quantizing Darboux coordinates on coadjoint orbit of Poincare group; obviously, they do commute. The price to pay is that they transform nonlinearly under rotations which linearize on $SO(2)$ subgroup only.  Some details are given in Appendix.

 \section{Newton-Wigner coordinate }

As we have seen in Sec.II the only case one can avoid the general conclusions of Mukunda and Skagerstam ( apart from the trivial case of scalar particles ) is $|\lambda|=1/2$ provided both helicities are taken into account. The dynamics (or rather kinematics) of such particles is described
by free massless Dirac equation:

\begin{equation}
\label{e17}
i\frac{\partial \Psi}{\partial t} = \vec{\alpha}\vec{p}  \Psi
\end{equation}

We work with the following representation of Dirac  matrices:

\begin{equation}
\label{e18}
\vec{\alpha}=
\left(
\begin{matrix}
\vec{\sigma} &0\\
0& - \vec{\sigma}
\end{matrix}
\right),
 \quad
\beta=\left(
\begin{matrix}
0 &I\\
I&0
\end{matrix} \right)
\end{equation}
The general positive energy solution to eq. (\ref{e17}) reads

\begin{equation}
\label{e19}
\Psi(x)=\frac{1}{(2\pi)^{3/2}}\int d^3\vec{p} \left[
\left( \begin{matrix}
u_+(p)\\
0
\end{matrix}
\right)
a_+(p)+
\left( \begin{matrix}
0\\
u_-(p)
\end{matrix}
\right)
a_-(p
\right]
e^{i\vec{p}\vec{x} - ip^0t}
\end{equation}
where $u_\pm(p)$ are spinors obeying

\begin{equation}
\label{e20}
\begin{split}
&p^k\sigma_k u_\pm(p)=\pm p^0u_\pm(p) \\
&u_\pm(p)^+u_\pm(p)=1, \quad u_\pm(p)^+u_\mp(p)=0
\end{split}
\end{equation}
One possible choice is

\begin{equation}
\label{e21}
\begin{split}
u_+(p)=\frac{1}{\sqrt{2p^0(p^0-p^3)}}
\left( \begin{matrix}
p^1-ip^2)\\
p^0-p^3
\end{matrix}
\right)  \\
u_-(p)=\frac{1}{\sqrt{2p^0(p^0+p^3)}}
\left( \begin{matrix}
-p^1+ip^2)\\
p^0+p^3
\end{matrix}
\right)
\end{split}
\end{equation}
The scalar product reads

\begin{equation}
\label{e22}
\int d^3 \vec{x} \Psi^+(\vec{x},t) \widetilde {\Psi}(\vec{x},t) = \int d^3 \vec{p} \left[ \overline{a}_+(p) \widetilde{a}_+(p) + \overline{a}_-(p) \widetilde{a}_-(p) \right]
\end{equation}

It differs from (\ref{e7}) by a simple redefinition of amplitudes and makes $i\partial_k$ (and $iD_k $)
hermitean at the expense of slightly complicating Poincare transformations rules.\\
The spaces of the amplitudes $ a_\pm(p) $ carry helicity $ \lambda=\pm\frac{1}{2} $ representations. The $ m=0 $ limit of Newton-Wigner operator \cite{b2}
, \cite{b3} reads

\begin{equation}
\label{e23}
X_k=x_k + \frac{i}{2} \left( \frac{\beta \alpha^k}{p^0}+\frac{(\vec{p}\vec{\alpha})\alpha^k-p^k}{(p^0)^2}- \frac{\beta(\vec{p}\vec{\alpha})p^k}{(p^0)^3}\right)
\end{equation}

with $ x_k $  and $ p^k $ having standard meaning.\\
Acting on the wave function (\ref{e19}) one finds that $ X_k $ preserves the positivity of energy, i.e. the result is of the same form (\ref{e19}) with new amplitudes $ a_\pm (p)$ given by

\begin{equation}
\label{e24}
\left(\begin{matrix}
a'_{+k}(p)\\
a'_{-k}(p)
\end{matrix}\right)=
\left[ i\partial_k + i
\left( \begin{matrix}
u^+_+ \partial_k u_+ & u^+_+ \partial_k u_-\\
u^+_- \partial_k u_+ & u^+_- \partial_k u_-
\end{matrix}\right)
\right]
 \left(\begin{matrix}
a_{+k}(p)\\
a_{-k}(p)
\end{matrix}\right)
\end{equation}
We see that the Newton-Wigner coordinate operator takes the form

\begin{equation}
\label{e25}
X_k=i{\mD}_k \equiv i \partial_k - {\mA}_k
\end{equation}

\begin{equation}
\label{e26}
 {\mA}_k \equiv (-i) \left( \begin{matrix}
u^+_+ \partial_k u_+ & u^+_+ \partial_k u_-\\
u^+_- \partial_k u_+ & u^+_- \partial_k u_-
\end{matrix}\right)
\end{equation}

and can be viewed as $ U(2) $ covariant derivative. \\

The non diagonal elements are nonvanishing so, as expected, both helicities are needed to define Newtom-Wigner coordinate. The diagonal elements are the monopole vector potentials describing monopoles of charges $ \pm 1/2 $ . They represent the potentials  corresponding to the Dirac strings extending along positive and negative third semiaxis, respectively. One can convert $ {\mA}_k $ into $ sU(2) $ element by extracting the trace which is a pure gauge except singularity along third axis. Then the diagonal elements of modified $ {\mA}_k $ describe monopoles in the gauge involving two strings along both positive and negative semiaxis, each carrying half of the total flux. However, for simplicity we will be dealing with $ U(2) $ connection given by (\ref{e26}).\\
The Newton-Wigner coordinates commute which implies that $ {\mA}_k $ should be a pure gauge. Indeed, it is immediately seen that
\begin{equation}
\label{e27}
 {\mA}_k =-i\mathcal{U}^+ \partial_k \mathcal{U}
\end{equation}
with
\begin{equation}
\label{e28}
 \mathcal{U}=\left( \begin{matrix}
 (u_+)_1 & (u_-)_1 \\
 (u_+)_2 & (u_+)_2
 \end{matrix}\right)
\end{equation}

Let us note that the nontrivial monopole connections describing irreducible representations of definite parity result from abelian connection (\ref{e27}) \cite{b17}; this is a counterpart of 'tHooft mechanism \cite{b18}.

\;\;
The gauge potential transforms according to the rule $ {\mA}'_k=V{\mA}_k V^+ -iV\partial_kV^+ $.
Choosing $ V=\mathcal{U}^+ $ we find $ {\mA}'_k=0 $. Therefore, in the new basis the coordinate operator
$ \hat{X}_k $ equals simply $ i\partial_k - \frac{i}{2}\frac{p^k}{(p^0)^2} $ (which is the partial derivative hermitean with respect to the scalar product (\ref{e7})). New generators of Lorentz group read
\begin{equation}
\label{e29}
\begin{split}
 &J_i= -i \varepsilon_{ijk} p^j \partial_k + \frac{1}{2} \sigma_i \\
 &K_i= i p^0 \partial_i + \frac{1}{2p^0} \varepsilon_{ijk}p^j \sigma_k
\end{split}
\end{equation}

 \section{The case $|\lambda|=1$ }

For completeness we discuss now the case $|\lambda|=1$ (photons). According to the arguments of Mukunda and Skagerstam, presented in sec. (II), for $ |\lambda|>1/2 $ it is not possible to find a coordinate operator obeying $(i) \div (iii) $, even if both helicities $ \pm |\lambda| $ are included. As an example, consider the case of photons, discussed already by Skagerstam \cite{b11}.\\
In the standard formalism \cite{b19} the photon wave function can be written as
\begin{equation}
\label{e30}
f_{\mu\nu}= \int \frac{d^3\vec{p}}{(2\pi)^32p^0}\left( e_{\mu\nu}(\vec{p})\varphi(\vec{p},1)+\overline{e_{\mu\nu}(\vec{p})}\varphi(\vec{p},-1)\right) e^{-ipx}
\end{equation}
with $ e_{\mu\nu}(\vec{p}) $ being the polarization tensor. Momentum amplitudes $ \varphi(\vec{p},\pm 1) $ span the massless representations of Poincare group corresponding to helicities $ |\lambda|=\pm 1 $. The scalar product in the space of states carrying both helicities (reducible representation ) reads
\begin{equation}
\label{e31}
 (\varphi,\psi)=\int \frac{d^3\vec{p}}{(2\pi)^32p^0}\left(\overline{\varphi(\vec{p},1)}\psi(\vec{p},1) + \overline{\varphi(\vec{p},-1)}\psi(\vec{p},-1)\right)
\end{equation}
Transformation rules for amplitudes $ \varphi(\vec{p},\lambda) $ read
\begin{equation}
\label{e32}
 \varphi(\vec{p},\lambda) = e^{i\lambda\Theta(p,\Lambda)}e^{i\vec{p}\vec{a}} \varphi(\overrightarrow{\Lambda^{-1}p},\lambda)
\end{equation}
The polarization tensor $  e_{\mu\nu}(\vec{p}) $ ( intertwinning operator ) can be expressed in terms of polarization vector $ \vec{e}(\vec {p}) $ \cite{b19}
\begin{equation}
\label{e33}
\begin{split}
 &e_{0i} = ip_0e_i(\vec{p})\\
 &e_{ij} = -p_0 \varepsilon_{ijk} e_k(\vec{p})
 \end{split}
\end{equation}

 obeying

 \begin{equation}
 \label{e34}
\begin{array}{ll}
\vec{p}\cdot\vec{e}(\vec{p})=0,\quad \quad & \vec{p}\times \vec{e}(\vec{p})= -i p_0 \vec{e}(\vec{p}) \\
 \vec{e}(\vec{p}) \cdot \vec{e}(\vec{p})=0,\quad \quad &  \overline{\vec{e}(\vec{p})} \cdot \vec{e}(\vec{p})=1 \\
\overline{ \vec{e}(\vec{p})}\times \vec{e}(\vec{p}) = i\frac{\vec{p}}{p_0} &
 \end{array}
\end{equation}

Let us define

 \begin{equation}
 \label{e35}
\vec{\varphi}(\vec{p})= \vec{e}(\vec{p})\varphi(\vec{p},1) + \overline{\vec{e}(\vec{p})}\varphi(\vec{p},-1)
\end{equation}

 Then

 \begin{equation}
 \label{e36}
\int \frac{d^3\vec{p}}{(2\pi)^32p^0}\overline{\vec{\varphi}(\vec{p})}\cdot\vec{\psi}(\vec{p}) = \int \frac{d^3\vec{p}}{(2\pi)^32p^0}\left(\overline{\varphi(\vec{p},1)}\psi(\vec{p},1) + \overline{\varphi(\vec{p},-1)}\psi(\vec{p},-1)\right)
\end{equation}

and

\begin{equation}
 \label{e37}
\vec{p}\cdot \vec{\varphi}(\vec{p}) = 0
\end{equation}

This is the form of representation discussed in Ref. \cite{b11}. The space of states is the transverse subspace of the Hilbert space of vector valued square integrable functions. The generators of the Poincare group consist of fourmomenta $ p^\mu $ together with the Lorentz group generators

\begin{equation}
 \label{e38}
\begin{split}
J_k& = \varepsilon_{klm}\left( i\partial_l - \frac{i}{2}\frac{p^l}{(p^0)^2}-{\mA}^a_l(\vec{p})\mathcal{S}_a\right)p^m \\
K_K &= \frac{1}{2}\left(p^0(i\partial_k-\frac{i}{2}\frac{p^k}{(p^0)^2}-{\mA}^a_k(\vec{p})\mathcal{S}_a) +
(i\partial_k-\frac{i}{2}\frac{p^k}{(p^0)^2}-{\mA}^a_k(\vec{p})\mathcal{S}_a)p^0\right)\\
& =  ip^0\partial_k -ip^0{\mA}^a_k(\vec{p})\mathcal{S}_a
\end{split}
\end{equation}

where

\begin{equation}
 \label{e39}
(\mathcal{S}_a)_{kl}=-i\varepsilon_{akl}
\end{equation}

is the spin generator and

\begin{equation}
 \label{e40}
{\mA}^a_k(\vec{p})=\varepsilon_{alk}\frac{p^l}{(p^0)^2}
\end{equation}

is the Wu-Yang monopole \cite{b20}. Note that $ {\mA}^a_k(\vec{p}) $ is regular on $ {\mR}^3 \smallsetminus \{ 0 \} $; this is related to the fact that principal $ SU(2) $ bundles over $ S^2 $ are trivial.\\
We see that again the generators are expressed in terms of covariant derivative ( this time corresponding to $ SU(2) $ connection):

\begin{equation}
 \label{e41}
({\mD}_k)_{lm}=i\delta_{lm}\partial_k - {\mA}^a_k(\mathcal{S}_a)_{lm}
\end{equation}

To make the contact with $ U(1) $ gauge symmetry described previously let us note that, by virtue of eqs. (\ref{e34}), the polarization vector $ \vec{e}(\vec{p}) $ can be written in the form \cite{b19}

\begin{equation}
 \label{e42}
\vec{e}(\vec{p})= \frac{1}{\sqrt{2}}\left(\vec{l}_1(\vec{p}) + i\vec{l}_2(\vec{p})\right)
\end{equation}

where $\vec{l}_1  $ and $ \vec{l}_2 $ obey

\begin{equation}
 \label{e43}
\vec{l}_1(\vec{p}) \times \vec{l}_2(\vec{p})=\frac{\vec{p}}{p^0}, \quad \quad \vec{l}_2(\vec{p}) \times \frac{\vec{p}}{p^0}= \vec{l}_1(\vec{p}),\quad \quad \frac{\vec{p}}{p^0} \times \vec{l}_1(\vec{p}) = \vec{l}_2(\vec{p})
\end{equation}

The set $  \{\vec{l}_1(\vec{p}), \vec{l}_2(\vec{p}) \}$ is defined up to an arbitrary rotation around $ \vec{p} $ by an  angle $ \omega (\vec{p}) $. This results in the following transformation of
$ \vec{e}(\vec{p}) $:

\begin{equation}
 \label{e44}
\vec{e}(\vec{p}) \rightarrow  e^{i\lambda\omega (\vec{p})}\vec{e}(\vec{p})
\end{equation}

which corresponds to a gauge transformation

\begin{equation}
 \label{e45}
\varphi(\vec{p},\lambda) \rightarrow  e^{-i\omega (\vec{p})}\varphi(\vec{p},\lambda)
\end{equation}

Again, as in the fermionic case, we are dealing with $ U(1) $ gauge symmetry corresponding to $ \lambda=\pm1 $ charges. In the original subspaces spanned by $ \varphi(\vec{p},\pm1) $ the Poincare generators are the functions of momenta and $ U(1) $ covariant derivatives

\begin{equation}
 \label{e46}
{\mD}_k^{\lambda} \equiv i\partial_k - {\mA}^{\lambda}_k
\end{equation}

where $ {\mA}^{\lambda}_k $ are the ( singular on $ {\mR}^3 \smallsetminus \{ 0 \} $ ) potentials of $ U(1) $ monopoles,

\begin{equation}
 \label{e47}
\partial_l {\mA}^{\lambda}_k-\partial_k {\mA}^{\lambda}_l = \lambda \varepsilon _{klm} \frac{p^m}{(p^0)^3}
\end{equation}

In order to find the relation between $ {\mD}_k $ and $ {\mD}_k^{\lambda} $ we write the consistency condition

\begin{equation}
 \label{e48}
i({\mD}_k)_{mn}\varphi_n(\vec{p}) = e_m(i\partial_k - {\mA}^{(+)}_k)\varphi(\vec{p},1) +
\overline{ e_m}(i\partial_k - {\mA}^{(-)}_k)\varphi(\vec{p},-1)
\end{equation}

which yields

\begin{equation}
 \label{e49}
 \begin{split}
{\mA}^{(+)}_k=-i\overline{ e_m}\partial_k e_m = \vec{l}_1\cdot\partial_k\vec{l}_2 \\
{\mA}^{(-)}_k=-ie_m \partial_k \overline{ e_m} = -\vec{l}_1\cdot\partial_k\vec{l}_2
\end{split}
\end{equation}

It is straightforward to check that eqs. (\ref{e49}) imply (\ref{e47}). We conclude that passing from the representation in terms of amplitudes $ \varphi(\vec{p},\pm1) $ to that based on $ \vec{\varphi}(\vec{p}) $
implies embedding the singular $ U(1) $ connection into regular $ SU(2) $ one.\\
Now, one can attempt to define the coordinate operator by

\begin{equation}
 \label{e50}
 \hat{X}_k = i {\mD}_k
\end{equation}

Then (i) and (ii) are fulfilled. However, the components of $  \hat{X}_k $ do not commute due to the nontriviality  monopole potential (\ref{e40}); instead one finds \cite{b11}

\begin{equation}
 \label{e51}
 [\hat{X}_k, \hat{X}_l] = -i \varepsilon_{klm} \frac{p^m}{(p^0)^4} p^a \mathcal{S}_a
\end{equation}

so (iii) is broken. This is due to lack of third helicity value, $ \lambda=0 $. Indeed, by virtue of

\begin{equation}
 \label{e52}
( \mathcal{S}_a\frac{p^a}{p^0})^2_{lm}=\delta_{lm} - \frac{p^lp^m}{(p^0)^2}
\end{equation}

one finds that the transversality condition (\ref{e37}) excludes $ \lambda=0 $. It is, however, the
transversality condition which makes both $ U(1) $ and $ SU(2) $ connections nontrivial. This is in contrast with $ \mid \lambda \mid = \frac{1}{2} $ case where the abelian projection yields nontrivial
 $ U(1) $ connection from trivial $ SU(2) $ one.

\section{Conclusions }

We have shown that, for the massless particles, $ \mid \lambda \mid = \frac{1}{2} $ is the only case ( apart from the trivial $ \mid \lambda \mid = 0 $ one )where the Newton-Wigner procedure works yielding the coordinate operator with commuting components. This is because taking massless Dirac equation one obtains reducible representation of Poincare group  comprising both helicities, $ \lambda=\pm \frac{1}{2} $.
Then the space of states carries twodimensional representation of $ SU(2) $ spin variables which, by the Mukunda-Skagerstam theorem, is a necessary condition for the existence of coordinate operator obeying natural conditions (i) $ \div $ (iii) listed in the Introduction. For higher helicities some states are lacking and the relevant $ SU(2) $ representation cannot be accommodated; the acceptable
( from the point of view (i) $ \div $ (iii) ) coordinate operator does not exist. From the technical point of view the lack of some helicities imposes the constraints on momentum amplitudes which makes impossible to embed the initial nontrivial $ U(1) $ connections into trivial nonabelian one. The appearance of covariant derivatives (and, consequently, the related connections ) is indispensable
for the representation of Poincare group to be independent of the convention used to define the standard state in the framework of induced representations.
\par
One can, of course, weaken the conditions   (i) $ \div $ (iii) resigning, for example, from explicit
rotation covariance ( the condition (ii) ). Then the coordinate operator can be constructed for any $ \lambda $, even within an irreducible representation. To this end one can start from Darboux coordinates on coadjoint orbit of Poincare group \cite{b16}. The resulting Hamiltonian system, upon quantization, yields the relevant irreducible representation of Poincare group. In particular, one obtains commuting coordinates. However, the price one has to pay is that they transform under rotation according to nonlinear realization linearizing  only on $ SO(2) $ subgroup of rotation group. An attempt to linearize the transformation rule leads to noncommuting coordinates. Some details are given in Appendix.
\par
Let us point out that the problem of chiral particles as classical massless spinning particles has been studied in the symplectic framework in a very interesting paper by Duval and Horvathy\cite{b21}
\section{Appendix }

\subsection{Nonminimal terms and rotational invariance}

As we have noted in Sec. III the correct form of rotation generators is obtained only after including an  additional term which is not generated by minimal coupling principle. Its appearance is related to the fact that there exists no rotationally invariant vector potential describing the magnetic monopole field. Here we discuss this issue in some details.
\par
A vector field transforms under rotations as follows

\begin{equation}
 \label{e53}
{\mA}'_i(p')=O_{ij}{\mA}_j(p), \quad \quad \quad p'^k=O_{kl}p^l
\end{equation}

where $O$ is orthogonal matrix representing rotation. Infinitesimally
$O_{ij}=\delta_{ij}+\omega_k\varepsilon_{kij}$ which yields

\begin{equation}
 \label{e54}
\delta{\mA}_i(p') \equiv {\mA}'_i(p) - {\mA}_i(p) = -\omega_k\varepsilon_{klj}p^j\partial_l{\mA}_i(p)
+\omega_k\varepsilon_{kij}{\mA}_j(p)
\end{equation}

For rotationally invariant field $ \delta{\mA}_i=0 $ and one finds

\begin{equation}
 \label{e55}
i\varepsilon_{klj}p^j\partial_l{\mA}_i(p) = [{\mA}_i(p), -i\varepsilon_{klj}p^l\partial_j]=i\varepsilon_{kij}{\mA}_j(p)
\end{equation}

In general, $ \delta{\mA}_i(p)\neq0 $. It is obvious that for vector potential generating rotationally invariant magnetic field is invariant up to a gauge transformation. In fact

\begin{equation}
 \label{e56}
\partial_i\delta {\mA}_j(p) - \partial_j\delta {\mA}_i(p) =
-\varepsilon_{ijm} (-\omega_k\varepsilon_{kln}p^n\partial_lB_m + \omega_k\varepsilon_{kml}B_l)
\end{equation}

which vanishes by virtue of eq.(\ref{e54}) if the field configuration $ B_l $ is rotationally invariant. Consequently, the change of vector potential can be compensated by a gauge transformation.
Let us note that eq. (\ref{e55}) may be rewritten in the form

\begin{equation}
 \label{e57}
 [{\mA}_i(p), \varepsilon_{klj}p^l(-{\mD}_j)]=i\varepsilon_{kij}{\mA}_j(p)
\end{equation}

where $ {\mD}_j $ is given by eq. (\ref{e12}). More generally

\begin{equation}
 \label{e58}
 \delta{\mA}_i(p)= \omega_k \left( i\varepsilon_{kij}{\mA}_j(p) + [{\mA}_i(p), \varepsilon_{klj}p^l(-{\mD}_j)] \right)
\end{equation}

Let us compute

\begin{equation}
 \label{e59}
 i\delta(i{\mD}_i)= \omega_k \left( i\varepsilon_{kij}(i{\mD}_j) + [i{\mD}_i, \varepsilon_{klj}p^l(-i{\mD}_j)] \right)
\end{equation}

One finds

\begin{equation}
 \label{e60}
 i\delta(i{\mD}_i)=i \omega_k \left( p^iB_k-\delta_{ik}p^lB_l \right)
\end{equation}

Using $ B_k=\frac{\lambda p^k}{(p^0)^3} $ eq. (\ref{e60}) becomes

\begin{equation}
 \label{e61}
 i\delta(i{\mD}_i)=i\lambda \omega_k \left( \frac{p^ip^k}{(p^0)^3}-\frac{\delta_{ik}}{p^0} \right)=
 -i\lambda \omega_k \partial_i\left(\frac{p^k}{p^0} \right)
\end{equation}

We see that in order to achieve invariance, $ \delta({\mD}_i)=0 $, one has to compensate rotation by a gauge transformation with gauge parameter $ \lambda \omega_k \frac{p^k}{p^0} $. This amounts to modification of rotation generator by adding $ \lambda \frac{p^k}{p^0} $.

\subsection{Generators in Wu-Yang embedding}

Consider the Skagerstam \cite{b11} description of photon states described in Sec. IV. The generators of the Lorentz group in the picture involving Wu-Yang connection read:

\begin{equation}
 \label{e62}
 \begin{split}
 (J_k)_{mn}=\varepsilon_{kij}(x_i)_{mn}p^j + \frac{p^k}{(p^0)^2}p^l(\mathcal{S}_l)_{mn}\\
 (K_k)_{mn}=\frac{1}{2}\left(p^0(x_k)_{mn}+(x_k)_{mn}p^0\right)
 \end{split}
\end{equation}

where

\begin{equation}
 \label{e63}
 \begin{split}
 &(x_k)_{mn}=\left( i\partial_k + \frac{i}{2} \frac{p^k}{(p^0)^2}\right)\delta_{mn}+
 i\frac{p^m}{(p^0)^2}\delta_{kn} - i\frac{p^n}{(p^0)^2}\delta_{km}\\
 &(\mathcal{S}_l)_{mn}= -i\varepsilon_{lmn}
 \end{split}
\end{equation}

 It is straightforward to check that the action Poincare generators preserves transversality condition $ p^k\varphi_k(p)=0 $.
 Let us first note that the rotation generators can be written in the known form

 \begin{equation}
 \label{e64}
 (J_k)_{mn}=\delta_{mn}\varepsilon_{kjl}p^j(-i\partial_l) + (\mathcal{S}_k)_{mn}
\end{equation}

On the other hand the boosts read

 \begin{equation}
 \label{e65}
 (K_k)_{mn}=i\delta_{mn}p^0\partial_k+i\frac{p^m}{(p^0)}\delta_{kn}-i\frac{p^n}{(p^0)}\delta_{km}
\end{equation}

\subsection{Nonlinearly transforming coordinates}

The coordinate operators obeying (i)$\div$(iii) can not be constructed unless $ \mid\lambda\mid=1/2 $
and both helicities are included. However, for any $ \lambda $ one can find coordinates obeying (i) and (ii); to this end it is sufficient to identify coordinate operator wih covariant derivative (multiplied by imaginary unit). It is also possible to construct the coordinate operator such that (i) and (iii) hold. The price one has to pay is that the transformation rules under rotations become more complicated: the coordinates transform nonlinearly and the action of rotation group linearizes on $ SO(2) $
subgroup of rotations around the direction of standard vector. The relevant construction is quite straightforward \cite{b16} . It is well known that the representations of the Poincare group can be obtained by quantization of the relevant Hamiltonian systems on coadjoint orbit. By Darboux theorem the symplectic form on coadjoint orbit can be put in standard form. The coordinate operators are then the quantum counterpart of Darboux coordinates. There is no point to repeat here the full reasoning \cite{b16}.
However, the final result is quite simple. The resulting coordinate operators differ from the ones obeying (i) and (ii) by a (minus) gauge potential in some fixed gauge. As a result they are basically the covariant derivatives corresponding to pure gauge configurations so they commute. However, the gauge field can at most have an explicit symmetry with respect to the rotations around one axis while the remaining rotations take nonlinear form (which is another way of expressing the property that the effect of rotations can be compensated by a gauge transformation).

\par
{\bf Acknowledgments}
 The author would like to thank Professors Krzysztof Andrzejewski, Joanna Gonera and Cezary Gonera for helpful discussions and useful remarks.\\
This work has been supported in part by the grant 2016/223/B/ST2/00727 of National Science Centre, Poland


TITLE: Definitions which should be propositions/theorems
QUESTION [7 upvotes]: I am asking for a list of concepts which some sources present as definitions whereas other sources pose them as propositions/theorems.
For example, most abstract algebra books will define a group isomorphism to be a bijective group homomorphism. However, after one is introduced to category theory, one realizes that it is an ever-so-slightly non-trivial result that isomorphisms in Grp are precisely the bijective group homomorphisms.
Another example is a $C^k$ differentiable manifold. It is a theorem that every maximal atlas of a $C^k$ differentiable manifold ($k>0$) contains a $C^\infty$ atlas. And thus, the nuances in the terms '$C^k$ differentiable manifold' and 'smooth manifold' are not discussed in some sources.
The final example I'll point out is analytic vs. holomorphic complex functions. I've seen books define holomorphic functions and then say 'analytic' is just a synonym. Whereas I believe that it should be a theorem that every holomorphic function is analytic (where analytic is of course defined to be 'representable by a convergent power series').
The problem with these examples is that without the proper background, I could live my life blissfully ignorant with using these terms as definitions. But I believe that this also robs me of seeing a beautiful result which hints at deep subtleties. So I am asking the community to share their knowledge of other such examples they may have encountered.

REPLY [2 votes]: I think that most introductions to the determinant are written in a very misleading way—they define the determinant using a formula or an algorithm, then gradually prove that it has various nice properties.  This makes it seem as though facts like $\operatorname{det}AB=(\operatorname{det}A)(\operatorname{det}B)$, $\operatorname{det}A=\operatorname{det}A^T$, or "$\operatorname{det}$ is linear in each row" are nontrivial, and makes the definition itself seem unmotivated.
Really, the one major nontrivial theorem about determinants is that they exist—that is, some function exists on the set of $n\times n$ matrices which is linear in each row, vanishes when two rows are equal, returns $1$ for the identity matrix, etc.  The various "definitions" of the determinants are easy to understand as consequences of these properties, but, pedagogically, they should appear after those properties are introduced, not before.
To further illustrate this example, consider the determinant when restricted to permutation matrices, which is just the sign function on permutations, $\epsilon: S_n \to \{\pm 1\}$ (and since I like stirring up controversy, I'll also mention that $\epsilon$ is the determinant on the set of $n\times n$ matrices over the field $\mathbb{F}_1$).
It is not obvious from the definitions that the sign of a permutation exists, i.e. that each $S_n$ ($n\geq 2$) has a normal subgroup of index $2$.  I'm not saying it's a difficult theorem, but there's definitely something that needs proving, namely: if a product of $k$ transpositions equals a product of $l$ transpositions, then $k$ and $l$ have the same parity.  Technically we can define $\epsilon(\sigma)$ to be $(-1)^{q_\sigma}$, where $q_\sigma$ is the number of pairs from $\{1,2,\ldots,n\}$ which are out of order, then use its properties to prove that $\epsilon$ is a homomorphism... but in my opinion, this definition is only motivated within the attempt to prove that there exists a homomorphism $\epsilon$ with certain properties.
(By contrast, something like the existence of "the" tensor product or "the" algebraic closure is usually handled correctly, in my experience—we give definitions to say what properties we want these things to have, then only give concrete forms for them when we need to prove that they exist.)


text:
\section{Ambiguous Case}
Tags: Triangles, Named Theorems

\begin{theorem}
Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
Let the [[Definition:Side of Polygon|sides]] $a, b, c$ of $\triangle ABC$ be [[Definition:Opposite (in Triangle)|opposite]] $A, B, C$ respectively.
Let the [[Definition:Side of Polygon|sides]] $a$ and $b$ be known.
Let the [[Definition:Angle|angle]] $\angle B$ also be known.
Then it may not be possible to know the value of $\angle A$.
This is known as the '''ambiguous case'''.
\end{theorem}

\begin{proof}
From the [[Law of Sines]], we have:
:$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$
from which:
:$\sin A = \dfrac {\sin a \sin B} {\sin b}$
We find that $0 < \sin A \le 1$.
We have that:
:$\sin A = \map \sin {\pi - A}$
and so unless $\sin A = 1$ and so $A = \dfrac \pi 2$, it is not possible to tell which of $A$ or $\pi - A$ provides the correct solution.
:[[File:Ambiguous-Case.png|400px]]
{{qed}}
\end{proof}

config: wiki


TITLE: Request for the proof of a theorem
QUESTION [1 upvotes]: I am sincerely requesting a proof of the following theorem from anyone willing to do so. We are about to learn about Sylow groups and simple groups next, and I want to get further ahead of the game by understanding this theorem because I have read ahead a little bit.
FYI, we are using Algebra by Artin.
Theorem: If G is a finite group with a subgroup H such that $|G|∤([G:H])!$ then G contains a non-trivial proper normal subgroup containing H.
Thanks!

REPLY [0 votes]: Here are the guides:
Define an action of $\;G\;$ on the set $\;X\;$ of left cosets of $\;H\;$ by
$$g\cdot(xH)\mapsto(gx)H$$
Prove the above indeed is an action (almost trivial).
Now, look at the induced homomorphism $\;\phi:G\to\text{ Sym}_X\cong S_n\;,\;\;n:=[G:H]\;$ ,  from the above action. Prove that $\;\ker\phi\;$ is the maximal normal subgroup of $\;G\;$ contained in $\;H\;$  . If you want, you can try to prove that in fact
$$\ker\phi=\bigcap_{g\in G}g^{-1}Hg=:\text{ the socle of}\;H$$
Apply now the first isomorphism theorem to $\;\phi\;$ and use the given info to reach the desired result.


TITLE: How do i define $f(z_0)$ in an contour integral?
QUESTION [0 upvotes]: I have a Problem described as following:
$$
\oint \dfrac{\frac{3}{7}}{z+3}+\dfrac{\frac{4}{7}}{z-4}\,\mathrm{dz} \;\;\text{ on the contour } |z-3|=2
$$
and i have come as far to use the formulars below, but i am not sure how to define $f(z_0)$
$$
f(z_0)=\dfrac{1}{2\pi i} \oint \dfrac{f(z)}{z-z_0},\mathrm{dz} = f(z_0)\cdot 2\pi i
$$
$z_0=4$ and $z_0=-3$ but only $z_0=4$ is inside the contour and is the only part i look at.
is $$f(z_0)=\frac{4}{7}\frac{1}{z-4}$$
if not how do i find it?
hope someone can help.

REPLY [1 votes]: Since the only point at which $\frac{3/7}{z+3}$ is undefined is outside the region bounded by the given contour, $\oint\frac{3/7}{z+3}\,\mathrm dz=0$. And so\begin{align}\oint\frac{3/7}{z+3}+\frac{4/7}{z-4}\,\mathrm dz&=\oint\frac{4/7}{z-4}\,\mathrm dz\\&=2\pi i\times\frac47\\&=\frac{8\pi i}7.\end{align}


TITLE: factorization of the cohomology of configuration space
QUESTION [6 upvotes]: This question is a follow-up to my previous question factorization of the regular representation of the symmetric group, which was answered in a very satisfactory way.
Let $\operatorname{Conf}(n,\mathbb{R}^3)$ be the configuration space of $n$ labeled points in $\mathbb{R}^3$, and consider the cohomology
$B_n := H^*(\operatorname{Conf}(n,\mathbb{R}^3))$, which is a graded representation of $S_n$.  Let $W_n := V[n] + q^2 V[n-1,1]$ be the graded representation that is the 1-dimensional trivial representation in degree 0 and the $(n-1)$-dimensional irreducible permutation representation in degree 2.
Question:  Does there exist a graded representation $M_n$ such that $M_n\otimes W_n\cong B_n$?
If we forget about the grading, then $B_n$ is just the regular representation and $W_n$ is the vector representation $\mathbb{C}^n$; the existence of a representation whose tensor product with the vector representation is isomorphic to the regular representation is explained in the aforementioned post.  So I am now asking a more refined version of the question in which the grading is being taken into account.
I will also note that I "know" that the answer is positive.  That is, I can define a graded representation $M_n$, I conjecture that there is an isomorphism $M_n\otimes W_n\cong B_n$,
and I've checked this conjecture on a computer up to $n=10$.  What I really want to know is whether the graded representation $B_n$ is already known to factor in this way.  If so, then I would really like to understand why this is the case, as I believe that it will help me to understand the representation $M_n$ that I am interested in.
(In case anyone would like to know, my representation $M_n$ that conjecturally solves this problem is the intersection cohomology of the hypertoric variety associated with the braid arrangement.)

REPLY [3 votes]: Here's a geometric construction of a factorization that works for points in $\mathbb R^2$ (and not any other dimension). Given the close relationship between the cohomology rings of configuration spaces of points in $\mathbb R^d$ for varying $d$ (cf my answer to Cohomology of configuration space as a representation of the symmetric group) I'd expect that one can extract a factorization in arbitrary dimension from this construction, even though the geometric part breaks down.
Note first that the affine group $G = \mathbb C \rtimes \mathbb C^\times$ is the subgroup of Möbius transformations fixing the point $\infty \in \mathbb P^1$. It follows that the quotient $\mathrm{Conf}(\mathbb C,n)/G$ is equal to the moduli space $M_{0,n+1}$ of $(n+1)$ points on the projective line modulo symmetries. This makes $\mathrm{Conf}(\mathbb C,n)$ homotopic to a trivial circle bundle over $M_{0,n+1}$, and so
$$ H^\bullet(\mathrm{Conf}(\mathbb C,n)) \cong H^\bullet(M_{0,n+1}) \oplus H^{\bullet-1}(M_{0,n+1}).$$
So one only needs to construct such an $S_n$-equivariant factorization for the cohomology $H^\bullet(M_{0,n+1})$. Now there is an $S_n$-equivariant  fiber bundle $M_{0,n+1} \to M_{0,n}$ by forgetting the last marking. Each fiber $F$ is $\mathbb P^1$ minus $n$ points, and so $H^0(F) \cong V[n]$, $H^1(F) \cong V[n-1,1]$ as $S_n$-representations. The Leray-Serre spectral sequence degenerates, and this gives the claimed factorization.


TITLE: Compute lattice points in rectangular region
QUESTION [2 upvotes]: Let $M \subset \mathbb R^2$ be a lattice of rank $2$ given by a basis $b_1,b_2\in \mathbb R^2$. So $M = \mathbb Z b_1 +\mathbb Z b_2$. I'm looking for an algorithm which computes every lattice point in a rectangular region $[a,b] \times [c,d]$. So we are interested in the finite set $M \cap ([a,b] \times [c,d])$.
In case $b_1 = (r_1,0)$ and $b_2 = (0,r_2)$ for $r_1,r_2 \in \mathbb R$ that would be quite simple. But my basis is of course not in that form (in fact for my lattices there is no basis of that form since $(0,0) \in M$ is the only lattice point with a zero in a coordinate).
Do you have ideas how to proceed? I guess I should transform my basis somehow into a basis which is more suitable. But what does more suitable mean in this situation and how can I do that?

REPLY [0 votes]: There are two relevant metrics in the problem, one is in ${\Bbb R}^2$ coming from the rectangle, the second in ${\Bbb Z}^2$ induced by your lattice basis.
In order to avoid carrying around on annoying constants (from the rectangle), the first reduction is to multiply $x,y$ coordinates by $1/(b-a)$ and $1/(d-c)$, respectively.
Your rectangle then becomes a unit square centered at $C=\left( \frac{a+b}{2(b-a)}, \frac{c+d}{2(d-c)}\right)$. For the lattice basis in these new coordinates we write:
$$ (b_1\;\; b_2) = (e_1\;\; e_2) \left( \begin{matrix} u_1 & v_{1} \\ u_2 & v_2 \end{matrix} \right)$$
with $e_1,e_2$ being the canonical basis in ${\Bbb R}^2$. The idea is now to perform change of lattice basis in order to make $b_1,b_2$ as 'orthogonal' as possible (with respect to the Euclidean metric, since we have rescaled the rectangle to a square). Thus we multiply on the right with integer matrices of determinant $\pm 1$ (to preserve the basis property) in order to reduce the angle between $b_1$ and $b_2$. A bit like Gram-Schmidt orthogonalization except that we have to stick to integer matrices.
A kind of Euclidean algorithm:
Order first $b_1$ and $b_2$ so that  $|b_1| \geq |b_2|$. Now, determine the integer $n$ that minimizes the length of
$$ b'_1 = b_1 - n b_2$$
Also set $b'_2=b_2$ and reverse the roles of $b_1$ and $b_2$, repeating the algorithm until $|b_1| |b_2|$ becomes smaller than or equal to $2|\det (b_1 \;\; b_2)|$. Note that the determinant is preserved under the algorithm and that the algorithm terminates. When terminated the angles between the resulting vectors $(b_1,b_2)$ is at least $60$ degrees.
It follows that (if I didn't make a mistake in the calculations):
$$ \frac{1}{4} \left( n_1^2 |b_1|^2 + n_2^2 |b_2|^2\right) \leq \left|n_1 b_1 + n_2 b_2\right|^2 \leq \frac{7}{4} \left( n_1^2 |b_1|^2 + n_2^2 |b_2|^2\right)$$
If centered at the origin the left hand inequality shows that is suffices to look at points with coordinates $|n_1|\leq 2/|b_1|$ and $|n_2|\leq 2/|b_2|$. I leave it to you the case when centered at a different point. Asymptotically, when both  $|b_1|$ and $|b_2|$ are much smaller than one, the right hand inequality shows that our method checks at most a factor of $\sqrt{7}$ too many points. The constant is probably not optimal, though.


TITLE: differentiability of $|xy|$ at $(0,0)$
QUESTION [0 upvotes]: Let, $f(x,y)=|xy|$, Show that $f$ is differentiable at $(0,0)$

I started with by trying to prove $\lim_{h\to 0,k\to 0}\dfrac{f(0+h,0+k)-f(0,0)}{hk}$ exists. Then, $\lim_{h\to 0,k\to 0}\dfrac{|hk|}{hk}=\lim_{h\to 0,k\to 0}\dfrac{|h||k|}{hk}=\lim_{h\to 0}\dfrac{|h|}{h}\lim_{k\to 0}\dfrac{|k|}{k}$
Can I do this? Because, if I do this, $f$ may not be differentiable, but I have to show the converse. Thanks, for any help!

REPLY [0 votes]: Your approach is not correct.
We have $ \frac{f(t,0)-f(0,0)}{t}=0 \to 0$ as $t \to 0$. Hence $f_x(0,0)$ exists and $=0$. The same arguments show that $f_y(0,0)$ exists and $=0$.
Now consider
$Q(x,y):= \frac{f(x,y)-f(0,0)-xf_x(0,0)-yf_y(0,0)}{\sqrt{x^2+y^2}}$.
Then we have: $f$ is differentiable at $(0,0) \iff Q(x,y) \to 0$ as $(x.y) \to (0,0)$.
It is your turn to show that $Q(x,y) \to 0$ as $(x.y) \to (0,0)$.


TITLE: Are there function with one local extremum but no global extrema?
QUESTION [4 upvotes]: Is it possible to find function $f : \mathbb R^2 \to \mathbb R$ such that $f$ is $C^2$ and has one, and only one local extremum but no global extremum.
I tried to make up an example with this function $f : (x,y) \mapsto e^{-(x^2+y^2)}+2x+2y$ whose graph looks like this :

I tried to show that $f$ has only one global maximum by showing there exists only one point where $\nabla f(x,y) = 0$ :
I obviously started by calculating, for $x, y \in \mathbb R$, $\nabla f (x,y)$:
$$\nabla f (x,y) = \begin{bmatrix} -2xe^{-(x^2+y^2)}+2\\ -2ye^{-(x^2+y^2) }+2 \end{bmatrix}$$
I need to show that the $\nabla f(x,y) = 0$ has only one solution and that it is a local extremum.
Any help would be apreciated!

REPLY [3 votes]: This is a classic question. The easiest example I know is
$$f(x,y) = x^3+e^{3y}-3xe^y.$$
I leave it to you to show that $f$ has precise one critical point, which is a local minimum. On the other hand, you can show easily that $f$ is unbounded (both ways).


TITLE: Integration in many variables.
QUESTION [2 upvotes]: Prove that
$$
\int_a^b\,\left(\int_a^y\,f\,(x,y)\,dx\right)\,dy\,=\,\int_a^b\,\left(\int_x^b\,f\,(x,y)\,dy\right)\,dx
$$

REPLY [1 votes]: Observe that $\chi_{[a,y]}(x) = [a \leq x \leq y]$.($\chi$ is the indicator function and $[\cdot]$ is the Iverson bracket) Then
$$\begin{align*} \int_a^b \int_a^y f(x,y)dx\,dy &= \int_a^b \int_a^b \chi_{[a,y]}(x) f(x,y)dx\,dy \\
&= \int_{[a,b]^2} \chi_{[a,y]}(x) f(x,y)dx \times y \\
&= \int_a^b \int_a^b [a \leq x \leq y] f(x,y)dy\,dx \\
&= \int_a^b \int_x^b f(x,y)dy\,dx
\end{align*}$$
I assume that $f$ satisfy the condition for Fubini-Tonelli Theorem.