text:
\begin{document}
\maketitle

\begin{abstract}
We give a detailed proof of the existence of evaluation map for affine Yangians of type A to clarify that it needs an assumption on parameters.
This map was first found by Guay but a proof of its well-definedness and the assumption have not been written down in the literature.
We also determine the highest weights of evaluation modules defined as the pull-back of integrable highest weight modules of the affine Lie algebra $\hat{\gl}_N$ by the evaluation map.
\end{abstract}

\section{Introduction}

The affine Yangian $\affY$ of type A is a two-parameter deformation of the universal enveloping algebra of the universal central extension of a double loop Lie algebra $\mathfrak{sl}_N[s,t^{\pm 1}]$.
It may be regarded as an additive degeneration of the quantum toroidal algebra.
While the representation theory of quantum toroidal algebras and affine Yangians is expected to be very rich, it is still mysterious and intricate.
The author constructed certain representations of $\affY$ in \cite{kodera_prims, kodera_TG}.
These results are degenerate analogs of works by Varagnolo-Vasserot~\cite{MR1626481}, Saito-Takemura-Uglov~\cite{MR1603798}, Takemura-Uglov~\cite{MR1710750}, Nagao~\cite{MR2583334} in the quantum toroidal case.
Other known constructions of representations are by the Schur-Weyl type functor \cite{MR2199856}, by geometric approaches \cite{MR1818101,MR2827177}, and by vertex operators \cite{GRW}.
A different method to construct representations of the affine Yangian is considered in this paper.

Guay introduced an algebra homomorphism from the affine Yangian $\affY$ to a completion of $U(\hat{\gl}_N)$ in \cite{MR2323534}.
This is an affine analog of the well-known evaluation map from the Yangian $Y(\mathfrak{sl}_N)$ to $U(\gl_N)$.
We refer to Guay's map as evaluation map for the affine Yangian.
Since the classical evaluation map for $Y(\mathfrak{sl}_N)$ plays a fundamental role for the construction of irreducible representations, Guay's evaluation map should be important to develop the representation theory of the affine Yangian.

In \cite{MR2323534}, explicit values of the evaluation map are given only for part of the generators of $\affY$.
It is stated that the values for other generators are determined from the defining relations of $\affY$ and a property of the map, and a proof of its well-definedness is omitted.
One of the goals of the present paper is to provide a detailed proof of the existence of the evaluation map for $\affY$.
To accomplish this, we give explicit values of the evaluation map for distinguished generators of degree one.
By results of Guay~\cite{MR2323534} and Guay-Nakajima-Wendlandt~\cite{GNW}, the affine Yangian is generated by these and degree zero elements with reasonable defining relations.

In the course of checking the well-definedness, we have realized that we need to impose a certain relation among the parameters and the central element of the affine Yangian.
To clarify this fact is also a purpose to write this paper.
So far we have not yet been able to define the evaluation map for general parameters.
The main results of the paper are Theorem~\ref{thm:Guay} and Theorem~\ref{thm:main}.
Theorem~\ref{thm:Guay} is Guay's original version.
The original Guay's evaluation map matches with lowest weight modules of $\hat{\gl}_N$.
To deal with highest weight modules, we introduce an opposite evaluation map in Theorem~\ref{thm:main}.

We can pull back integrable highest weight modules of $\hat{\gl}_N$ via the evaluation map in Theorem~\ref{thm:main} to make them $\affY$-modules.
The resulting modules satisfy the highest weight condition for affine Yangian.
We determine the highest weights of these evaluation modules in Theorem~\ref{thm:highest_weight} and investigate an analog of the Drinfeld polynomials.
The level-one Fock representation constructed by the author in \cite{kodera_prims, kodera_TG} turns out to be isomorphic to an evaluation module if we specialize the parameters.

The quantum toroidal case is studied by Miki~\cite{MR1694256} and Feigin-Jimbo-Mukhin~\cite{FJM}.
However, the evaluation map we consider here is not a direct analog of theirs.
It seems to be a degeneration of the homomorphism in \cite{MR1694256, FJM} twisted by Miki's automorphism.

This paper is organized as follows.
Section~2 is devoted to preliminaries on the affine Yangian $\affY$ and the affine Lie algebra $\hat{\gl}_N$.
We provide completions of $U(\hat{\gl}_N)$ which are used to formulate the main results.
In Section~3, we give a proof of the existence of the two kinds of evaluation map.
Then we determine the highest weights of evaluation modules in Section~4.

\subsection*{Acknowledgments}
This work was supported by JSPS KAKENHI Grant Number 17H06127 and 18K13390.

\section{Preliminaries}\label{sec:affine_Yangian}

\subsection{Affine Yangian}

Fix an integer $N \geq 3$ throughout the paper.
We use the notation $\{x,y\}=xy+yx$.

\begin{dfn}
The affine Yangian $\affY = Y_{\ve_1,\ve_2}(\affsl)$ is the algebra over $\bbC$ generated by $x_{i,r}^{+}, x_{i,r}^{-}, h_{i,r}$ $(i \in \mathbb{Z} / N\mathbb{Z}, r \in \mathbb{Z}_{\geq 0})$ with parameters $\ve_1, \ve_2 \in \bbC$ subject to the relations:
\begin{equation*}
	[h_{i,r}, h_{j,s}] = 0,
	\quad [x_{i,r}^{+}, x_{j,s}^{-}] = \delta_{ij} h_{i, r+s},
	\quad [h_{i,0}, x_{j,r}^{\pm}] = \pm a_{ij} x_{j,r}^{\pm},
\end{equation*}
\begin{equation*}
	[h_{i, r+1}, x_{j, s}^{\pm}] - [h_{i, r}, x_{j, s+1}^{\pm}]
	= \pm a_{ij} \dfrac{\varepsilon_1 + \varepsilon_2}{2} \{h_{i, r}, x_{j, s}^{\pm}\}
	- m_{ij} \dfrac{\varepsilon_1 - \varepsilon_2}{2} [h_{i, r}, x_{j, s}^{\pm}],
\end{equation*}
\begin{equation*}
	[x_{i, r+1}^{\pm}, x_{j, s}^{\pm}] - [x_{i, r}^{\pm}, x_{j, s+1}^{\pm}]
	= \pm a_{ij}\dfrac{\varepsilon_1 + \varepsilon_2}{2} \{x_{i, r}^{\pm}, x_{j, s}^{\pm}\}
	- m_{ij} \dfrac{\varepsilon_1 - \varepsilon_2}{2} [x_{i, r}^{\pm}, x_{j, s}^{\pm}],
\end{equation*}
\begin{equation*}
	\sum_{w \in \mathfrak{S}_{1 - a_{ij}}}[x_{i,r_{w(1)}}^{\pm}, [x_{i,r_{w(2)}}^{\pm}, \dots, [x_{i,r_{w(1 - a_{ij})}}^{\pm}, x_{j,s}^{\pm}]\dots]] = 0 \ \ (i \neq j),
\end{equation*}
where
\[
	a_{ij} =
	\begin{cases}
		2  &\text{if } i=j, \\
		-1 &\text{if } i=j \pm 1, \\
		0  &\text{otherwise,}
	\end{cases}\quad
	m_{ij} =
	\begin{cases}
		1  &\text{if } j=i-1, \\
		-1 &\text{if } j=i+1, \\
		0  &\text{otherwise.}
	\end{cases}
\]
\end{dfn}
The subalgebra generated by $x_{i,0}^+, x_{i,0}^-, h_{i,0}$ ($i \in \bbZ / N\bbZ$) is isomorphic to $U(\affsl)$ (see \cite[Section~2]{GNW}).
We set $\hbar = \ve_1 + \ve_2$ and $\tilde{h}_{i,1} = h_{i,1} - \dfrac{\hbar}{2} h_{i,0}^2$.

\begin{thm}[Guay~\cite{MR2323534}, Proposition~2.1, Guay-Nakajima-Wendlandt~\cite{GNW}, Theorem~2.12 and Section~6]\label{thm:GNW}
The affine Yangian $\affY$ is isomorphic to the algebra generated by $x_{i,r}^{+}, x_{i,r}^{-}, h_{i,r}$ $(i \in \mathbb{Z} / N\mathbb{Z}, r = 0,1)$ subject to the relations:
\begin{equation*}
[h_{i,r}, h_{j,s}] = 0,
\end{equation*}
\begin{equation*}
[x_{i,0}^{+}, x_{j,0}^{-}] = \delta_{ij} h_{i, 0},
 \quad [x_{i,1}^{+}, x_{j,0}^{-}] = \delta_{ij} h_{i, 1} = [x_{i,0}^{+}, x_{j,1}^{-}],
\end{equation*}
\begin{equation*}
\ [h_{i,0}, x_{j,r}^{\pm}] = \pm a_{ij} x_{j,r}^{\pm},
 \quad [\tilde{h}_{i,1}, x_{j,0}^{\pm}] = \pm a_{ij}\left(x_{j,1}^{\pm}-m_{ij}\dfrac{\varepsilon_1 - \varepsilon_2}{2} x_{j, 0}^{\pm}\right),
\end{equation*}
\begin{equation*}
[x_{i, 1}^{\pm}, x_{j, 0}^{\pm}] - [x_{i, 0}^{\pm}, x_{j, 1}^{\pm}] = \pm a_{ij}\dfrac{\varepsilon_1 + \varepsilon_2}{2} \{x_{i, 0}^{\pm}, x_{j, 0}^{\pm}\} - m_{ij} \dfrac{\varepsilon_1 - \varepsilon_2}{2} [x_{i, 0}^{\pm}, x_{j, 0}^{\pm}],
\end{equation*}
\begin{equation*}
(\ad x_{i,0}^{\pm})^{1-a_{ij}} (x_{j,0}^{\pm})= 0 \ \ (i \neq j), \label{eq:16}
\end{equation*}
where we set $\tilde{h}_{i,1} = {h}_{i,1} - \dfrac{\ve_1 + \ve_2}{2} h_{i,0}^2$.
\end{thm}

We can slightly reduce the relations as follows.
Proofs are straightforward.

\begin{lem}\label{lem:reduce1}
The relation $[h_{i,0}, x_{j,1}^{\pm}]=\pm a_{ij} x_{j,1}^{\pm}$ is deduced from the following:
\[
	[\tilde{h}_{i,1}, h_{j,0}] = 0,\quad [h_{i,0}, x_{j,0}^{\pm}] = \pm a_{ij} x_{j,0}^{\pm},\quad [\tilde{h}_{i,1}, x_{i,0}^{\pm}]= \pm 2 x_{i,1}^{\pm}.
\]
\end{lem}

\begin{lem}\label{lem:reduce2}
The relation $[x_{i,0}^{+}, x_{j,1}^{-}]=\delta_{ij} h_{i, 1}$ is deduced from the following:
\[
	[\tilde{h}_{i,1}, h_{j,0}] = 0,\quad [x_{i,r}^{+}, x_{j,0}^{-}]=\delta_{ij} h_{i, r} \ \ \text{for $r=0,1$},
\]
\[ [\tilde{h}_{i,1}, x_{j,0}^{+}]= a_{ij}\left(x_{j,1}^{+}-m_{ij}\dfrac{\varepsilon_1 - \varepsilon_2}{2} x_{j, 0}^{+}\right),\quad	[\tilde{h}_{i,1}, x_{i,0}^{-}] = -2 x_{i,1}^{-}.
\]
\end{lem}

For each $\alpha \in \bbC$, we define an algebra automorphism $\tau_{\alpha}$ of $\affY$ by
\[
	x_{i,r}^{\pm} \mapsto \sum_{s=0}^r \dbinom{r}{s} \alpha^{r-s} x_{i,s}^{\pm}, \quad h_{i,r} \mapsto \sum_{s=0}^r \dbinom{r}{s} \alpha^{r-s} h_{i,s}.
\]
We can verify that $\tau_{\alpha}$ is well-defined in the same way as an automorphism $\rho$ given in \cite[Lemma~3.5]{MR2199856}.
The definition of $\rho$ will be recalled in \ref{subsection:cyclic}.
Let $\mu$ be the algebra anti-isomorphism from $Y_{\ve_1,\ve_2}(\affsl)$ to $Y_{-\ve_2,-\ve_1}(\affsl)$ defined by
\[
	x_{i,r}^{\pm} \mapsto - x_{i,r}^{\pm}, \quad h_{i,r} \mapsto - h_{i,r}.
\]
It is easy to see that the assignment respects the defining relations.

\subsection{Affine Lie algebra $\hat{\gl}_N$}\label{subsection:affine_Lie_algebra}

Let $\gl_N$ be the complex general linear Lie algebra consisting of $N \times N$ matrices.
We denote by $E_{i,j}$ the matrix unit with $(i,j)$-th entry $1$.
The indices $i,j$ of $E_{i,j}$ are regarded as elements of $\bbZ/N\bbZ$.
The transpose of an element $X$ of $\gl_N$ is denoted by ${}^t X$.

Let $\hat{\gl}_N = \gl_N \otimes \bbC[t,t^{-1}] \oplus \bbC c$ be the affine Lie algebra whose Lie bracket is given by
\[
	[X \otimes t^{r}, Y \otimes t^{s}] = [X,Y] \otimes t^{r+s} + r \delta_{r+s,0} \tr(XY) c, \quad \text{$c$ is central}.
\]
We denote the element $X \otimes t^s$ by $X(s)$.
We set
\begin{gather*}
	x_0^+ = E_{N,1}(1), \quad x_0^{-} = E_{1,N}(-1), \quad h_0 = E_{N,N} - E_{1,1} + c,\\
	x_{i}^{+} = E_{i,i+1}, \quad x_i^{-} = E_{i+1,i}, \quad h_i = E_{i,i}-E_{i+1,i+1} \ \ (i \neq 0),\quad \bfid = \sum_{i=1}^N E_{i,i}.
\end{gather*}
Let $\affnil_{\pm}$ be the Lie subalgebras of $\hat{\gl}_N$ generated by $x_{i}^{\pm}$ ($i \in \bbZ/N\bbZ$) and $\bfid(s)$ ($\pm s > 0$).
That is,
\[
	\affnil_{+} = \bigoplus_{\substack{i < j\\ s \geq 0}} \bbC E_{i,j}(s) \oplus \bigoplus_{\substack{i \geq j\\ s > 0}} \bbC E_{i,j}(s), \quad \affnil_{-} = \bigoplus_{\substack{i > j\\ s \leq 0}} \bbC E_{i,j}(s) \oplus \bigoplus_{\substack{i \leq j\\ s < 0}} \bbC E_{i,j}(s).
\]
Let $\affCar$ be the Cartan subalgebra generated by $h_i$ ($i \in \bbZ/N\bbZ$) and $\bfid$.

Let $\omega_U$ be the algebra anti-automorphism of ${U}(\hat{\gl}_N)$ defined by $\omega_U(X(s))= {}^{t}X(-s)$ and $\omega_U(c)=c$.
We denote by $\mu_U$ the algebra anti-automorphism of ${U}(\hat{\gl}_N)$ induced from $X \mapsto -X$ ($X \in \hat{\gl}_N$).
The anti-isomorphism $\mu$ defined in the previous subsection is an extension of the restriction of $\mu_U$ to $\affsl$.

Let us introduce two completions of $U(\hat{\gl}_N)$.
We define gradings of $\affnil_{\pm}$ by $\deg X(s) = s$.
Then $U(\affnil_{\pm})$ become graded algebras.
We denote by $U(\affnil_{\pm})[s]$ the degree $s$ components.
\begin{dfn}
We put
\[
	U(\hat{\gl}_N)_{{\rm comp},+} = \bigoplus_{k \in \bbZ}\, \prod_{\substack{r, s \geq 0\\ s-r = k}} \left( U(\affnil_-)[-r] \otimes U(\affCar) \otimes U(\affnil_+)[s] \right)
\]
and
\[
	U(\hat{\gl}_N)_{{\rm comp},-} = \bigoplus_{k \in \bbZ}\, \prod_{\substack{r, s \geq 0\\ r-s = k}} \left( U(\affnil_+)[r] \otimes U(\affCar) \otimes U(\affnil_-)[-s] \right).
\]
\end{dfn}
The both $U(\hat{\gl}_N)_{{\rm comp},+}$ and $U(\hat{\gl}_N)_{{\rm comp},-}$ have natural algebra structures which contain $U(\hat{\gl}_N)$ as subalgebras.
Moreover the algebra automorphism $\omega_U$ of $U(\hat{\gl}_N)$ extends to both and the algebra anti-automorphism $\mu_U$ of $U(\hat{\gl}_N)$ extends to an algebra anti-isomorphism
\[
	\mu_U \colon U(\hat{\gl}_N)_{{\rm comp},+} \to U(\hat{\gl}_N)_{{\rm comp},-}.
\]

\section{Evaluation map}
\subsection{Main theorem}
From now on, we evaluate the central element $c$ at a complex number, which is denoted also by the same letter $c$.

\begin{thm}[Guay~\cite{MR2323534}]\label{thm:Guay}
Assume $\hbar c = N \ve_2$.
Then there exists an algebra homomorphism $\ev \colon \affY \to U(\hat{\gl}_N)_{{\rm comp},-}$ uniquely determined by
\begin{gather*}
	\ev(x_{i,0}^{+}) = x_{i}^{+}, \quad \ev(x_{i,0}^{-}) = x_{i}^{-},\quad \ev(h_{i,0}) = h_{i},
\end{gather*}
\begin{gather*}
	\ev(x_{i,1}^{+}) = \begin{cases}
		(1+ N \varepsilon_2) x_{0}^{+} + \hbar \displaystyle\sum_{s \geq 0} \sum_{k=1}^N E_{k,1}(s+1) E_{N,k}(-s) \text{ if $i = 0$},\\
		(1+ i \varepsilon_2) x_{i}^{+} + \hbar  \displaystyle\sum_{s \geq 0} \Big( \sum_{k=1}^i E_{k,i+1}(s) E_{i,k}(-s) + \sum_{k=i+1}^N E_{k,i+1}(s+1) E_{i,k}(-s-1) \Big) \\
		\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{ if $i \neq 0$},
	\end{cases}
\end{gather*}
\begin{gather*}
	\ev(x_{i,1}^-) = \begin{cases}
		(1+ N \varepsilon_2) x_{0}^{-} + \hbar \displaystyle\sum_{s \geq 0} \sum_{k=1}^N E_{k,N}(s) E_{1,k}(-s-1) \text{ if $i = 0$},\\
		(1+ i \varepsilon_2) x_{i}^{-} + \hbar \displaystyle\sum_{s \geq 0} \Big( \sum_{k=1}^i E_{k,i}(s) E_{i+1,k}(-s) + \sum_{k=i+1}^N E_{k,i}(s+1) E_{i+1,k}(-s-1) \Big) \\
		\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{ if $i \neq 0$},
	\end{cases}
\end{gather*}
\begin{gather*}
	\ev(h_{i,1}) = \begin{cases}
		(1+ N \varepsilon_2) h_{0} - \hbar E_{N,N} (E_{1,1}-c) \\
		\quad + \hbar \displaystyle\sum_{s \geq 0} \sum_{k=1}^{N} \Big( E_{k,N}(s) E_{N,k}(-s) - E_{k,1}(s+1) E_{1,k}(-s-1) \Big) \text{ if $i = 0$},\\
		\\
		(1+ i \varepsilon_2) h_{i} - \hbar E_{i,i}E_{i+1,i+1} \\
		\quad+ \hbar \displaystyle\sum_{s \geq 0} \Big( \sum_{k=1}^{i} E_{k,i}(s) E_{i,k}(-s) + \displaystyle\sum_{k=i+1}^{N}  E_{k,i}(s+1) E_{i,k}(-s-1) \\
		\qquad\qquad\quad - \displaystyle\sum_{k=1}^{i}E_{k,i+1}(s) E_{i+1,k}(-s) - \displaystyle\sum_{k=i+1}^{N} E_{k,i+1}(s+1) E_{i+1,k}(-s-1) \Big)\\
		\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{ if $i \neq 0$}.
	\end{cases}
\end{gather*}
\end{thm}
The formulas in the theorem are deduced from those for $\ev(H_{i,1})$ $(i \neq 0)$ where $H_{i,1}=h_{i,1} + (i/2)(\ve_1-\ve_2)h_{i,0}$, given in \cite[Section~6, pp.\ 462--463]{MR2323534}.
In \cite{MR2323534}, computations for the well-definedness are omitted.
Moreover we need the condition $\hbar c=N\ve_2$ to verify that those formulas preserve the defining relations of $\affY$ and this condition is not mentioned in the paper.
By these reasons, we give a detailed proof of the existence of the evaluation map $\ev$ as one of the main results of the present paper.

\begin{rem}
It is straightforward to deduce the following formula for $\ev(\tilde{h}_{i,1})$:
\begin{equation*}
	\begin{split}
	\ev(\tilde{h}_{0,1}) &= (1+ N \varepsilon_2) h_{0} - \dfrac{\hbar}{2} (E_{N,N}^2 +  (E_{1,1}-c)^2) \\
		&\qquad+ \hbar \sum_{s \geq 0} \sum_{k=1}^{N} \Big( E_{k,N}(s) E_{N,k}(-s) - E_{k,1}(s+1) E_{1,k}(-s-1) \Big),
	\end{split}
\end{equation*}
\begin{equation*}
	\begin{split}
	&\ev(\tilde{h}_{i,1}) = (1+ i \varepsilon_2) h_{i} - \dfrac{\hbar}{2} (E_{i,i}^2+E_{i+1,i+1}^2) \\
		&\qquad+ \hbar \sum_{s \geq 0} \Big( \sum_{k=1}^{i} E_{k,i}(s) E_{i,k}(-s) + \displaystyle\sum_{k=i+1}^{N}  E_{k,i}(s+1) E_{i,k}(-s-1) \\
		&\qquad\qquad\qquad - \displaystyle\sum_{k=1}^{i}E_{k,i+1}(s) E_{i+1,k}(-s) - \displaystyle\sum_{k=i+1}^{N} E_{k,i+1}(s+1) E_{i+1,k}(-s-1) \Big)\ \ (i \neq 0).
	\end{split}
\end{equation*}
\end{rem}

\begin{proof}
By the definition of $\ev$, the elements $\ev(x_{i,0}^{\pm})$, $\ev(h_{i,0})$ ($i \in \bbZ/N\bbZ$) automatically satisfy the defining relations of $\affsl$.
By Theorem~\ref{thm:GNW}, Lemma~\ref{lem:reduce1} and \ref{lem:reduce2}, it is enough to show the following:

\begin{equation}
[\ev(h_{i,1}), h_{j}] = 0, \label{eq:mainthmHH1}
\end{equation}
\begin{equation}
[\ev(h_{i,1}), \ev(h_{j,1})] = 0, \label{eq:mainthmHH2}
\end{equation}
\begin{equation}
[\ev(x_{i,1}^{+}), x_{j}^{-}] = \delta_{ij} \ev(h_{i, 1}), \label{eq:mainthm+-}
\end{equation}
\begin{equation}
[\ev(\tilde{h}_{i,1}), x_{j}^{\pm}] = \pm a_{ij} \left( \ev(x_{j,1}^{\pm})-m_{ij}\dfrac{\varepsilon_1 - \varepsilon_2}{2} x_{j}^{\pm} \right), \label{eq:mainthmHX}
\end{equation}
\begin{equation}
[\ev(x_{i, 1}^{\pm}), x_{j}^{\pm}] - [x_{i}^{\pm}, \ev(x_{j, 1}^{\pm})] = \pm a_{ij}\dfrac{\hbar}{2} \{x_{i}^{\pm}, x_{j}^{\pm}\} - m_{ij} \dfrac{\varepsilon_1 - \varepsilon_2}{2} [x_{i}^{\pm}, x_{j}^{\pm}]. \label{eq:mainthmXX2}
\end{equation}
Moreover the relations (\ref{eq:mainthmHX}), (\ref{eq:mainthmXX2}) for $-$ are deduced from those for $+$ by applying the anti-automorphism $\omega_U$ since we have $\omega_U(\ev(h_{i,1}))=\ev(h_{i,1})$ and $\omega_U(\ev(x_{i,1}^+))=\ev(x_{i,1}^{-})$.

Let us start to check the relations.
We use the symbol $\delta(P)$ for $1$ if $P$ is true, $0$ otherwise.
The relation~(\ref{eq:mainthmHH1}) clearly holds since $\ev(h_{i,1})$ has only weight $0$ terms.

We show (\ref{eq:mainthmHH2}).
We may assume $i < j$.
Further assume $i \neq 0$.
The proof for the case $i=0$ is similar.
Put
\begin{gather*}
	A_i = \sum_{r \geq 0} \sum_{k=1}^i E_{k,i}(r) E_{i,k}(-r), \quad	B_i = \sum_{r \geq 0} \sum_{k=i+1}^N E_{k,i}(r+1) E_{i,k}(-r-1), \\
	C_i = \sum_{r \geq 0} \sum_{k=1}^i E_{k,i+1}(r) E_{i+1,k}(-r), \quad D_i = \sum_{r \geq 0} \sum_{k=i+1}^N E_{k,i+1}(r+1) E_{i+1,k}(-r-1).
\end{gather*}
Since we have $[\ev(h_{i,1}),E_{k,k}]=0$ for all $k$, it is enough to show
\[
	[A_i+B_i-C_i-D_i,\, A_j+B_j-C_j-D_j] = 0.
\]
We give a proof for
\begin{equation}
	[A_i,A_j] + [A_i,B_j] + [B_i,A_j] + [B_i,B_j] = 0. \label{eq:vanish}
\end{equation}
We can similarly prove
\begin{gather*}
	[A_i,C_j] + [A_i,D_j] + [B_i,C_j] + [B_i,D_j] = 0,\\
	[C_i,A_j] + [C_i,B_j] + [D_i,A_j] + [D_i,B_j] = 0,\\
	[C_i,C_j] + [C_i,D_j] + [D_i,C_j] + [D_i,D_j] = 0.
\end{gather*}
The identity (\ref{eq:vanish}) follows from next lemma.

\begin{lem}
We have
\[
	[A_i,A_j] = \displaystyle\sum_{\substack{r,s \geq 0\\r>s}} \sum_{k=1}^i \Big( E_{k,i}(r)E_{i,j}(s-r)E_{j,k}(-s) - E_{k,j}(s)E_{j,i}(r-s)E_{i,k}(-r) \Big), \quad [A_i,B_j] = 0,
\]
\begin{equation*}
	\begin{split}
		&[B_i,A_j] \\
		&= \sum_{r,s \geq 0} \Bigg( \sum_{k=1}^i \Big(\! -E_{k,i}(r+s+1)E_{i,j}(-r-1)E_{j,k}(-s) + E_{k,j}(s)E_{j,i}(r+1)E_{i,k}(-r-s-1) \Big) \\
		&\qquad\qquad + \sum_{k=j+1}^N \Big(\! -E_{k,i}(r+1)E_{i,j}(s)E_{j,k}(-r-s-1) + E_{k,j}(r+s+1)E_{j,i}(-s)E_{i,k}(-r-1) \Big) \Bigg),
	\end{split}
\end{equation*}
\begin{equation*}
	\begin{split}
		&[B_i,B_j] \\
		&= \displaystyle\sum_{\substack{r,s \geq 0\\r \leq s}} \sum_{k=j+1}^N \Big( E_{k,i}(r+1)E_{i,j}(s-r)E_{j,k}(-s-1) - E_{k,j}(s+1)E_{j,i}(r-s)E_{i,k}(-r-1) \Big).
	\end{split}
\end{equation*}
\end{lem}
\begin{proof}
We compute $[A_i,A_j]$ as
\begin{equation*}
	\begin{split}
		&[A_i,A_j] = \sum_{r,s \geq 0} \sum_{\substack{1 \leq k \leq i\\ 1 \leq l \leq j}}[E_{k,i}(r) E_{i,k}(-r),E_{l,j}(s) E_{j,l}(-s)]\\
		&= \sum_{r,s \geq 0} \sum_{\substack{1 \leq k \leq i\\ 1 \leq l \leq j}} \Big( [E_{k,i}(r),E_{l,j}(s)]E_{i,k}(-r)E_{j,l}(-s) + E_{k,i}(r) [E_{i,k}(-r),E_{l,j}(s)]E_{j,l}(-s)\\
		&\qquad\qquad\qquad\qquad + E_{l,j}(s)[E_{k,i}(r),E_{j,l}(-s)]E_{i,k}(-r) + E_{l,j}(s)E_{k,i}(r)[E_{i,k}(-r),E_{j,l}(-s)] \Big)\\
		&= \sum_{r,s \geq 0}\sum_{k=1}^i \Big( E_{k,j}(r+s)E_{i,k}(-r)E_{j,i}(-s) + E_{k,i}(r)E_{i,j}(s-r)E_{j,k}(-s) \\
		&\qquad\qquad\qquad\qquad - E_{k,j}(s)E_{j,i}(r-s)E_{i,k}(-r) - E_{i,j}(s)E_{k,i}(r)E_{j,k}(-r-s) \Big).
	\end{split}
\end{equation*}
In the last equality, we change the index $l$ to $k$.
The sum of the terms containing
\[
	E_{k,i}(a)E_{i,j}(b)E_{j,k}(c) \ (a,b,c \in \bbZ)
\]
is
\begin{equation*}
	\begin{split}
		&\sum_{r,s \geq 0}\sum_{k=1}^i \Big( E_{k,i}(r)E_{i,j}(s-r)E_{j,k}(-s) - E_{k,i}(r)E_{i,j}(s)E_{j,k}(-r-s) + E_{k,j}(r+s)E_{j,k}(-r-s)\Big) \\
		&= \sum_{\substack{r,s \geq 0\\r>s}}\sum_{k=1}^i E_{k,i}(r)E_{i,j}(s-r)E_{j,k}(-s) + \sum_{r,s \geq 0}\sum_{k=1}^i E_{k,j}(r+s)E_{j,k}(-r-s).
	\end{split}
\end{equation*}
Similarly the sum of the terms containing
\[
	E_{k,j}(a)E_{j,i}(b)E_{i,k}(c) \ (a,b,c \in \bbZ)
\]
is
\begin{equation*}
	-\sum_{\substack{r,s \geq 0\\r>s}}\sum_{k=1}^i E_{k,j}(s)E_{j,i}(r-s)E_{i,k}(-r) - \sum_{r,s \geq 0}\sum_{k=1}^i E_{k,j}(r+s)E_{j,k}(-r-s).
\end{equation*}
Hence the first identity holds.

The second identity $[A_i,B_j]=0$ is clear due to the condition $i < j$.

A direct computation shows
\begin{equation*}
	\begin{split}
		&[B_i,A_j] \\
		&= \sum_{r,s \geq 0} \Bigg( \sum_{k=i+1}^N E_{k,j}(r+s+1)E_{i,k}(-r-1)E_{j,i}(-s) - \sum_{k=1}^j E_{k,i}(r+s+1)E_{i,j}(-r-1)E_{j,k}(-s) \\
		&\qquad\qquad\quad + \sum_{k=i+1}^j \Big( E_{k,i}(r+1)E_{i,j}(s-r-1)E_{j,k}(-s) - E_{k,j}(s)E_{j,i}(r-s+1)E_{i,k}(-r-1) \Big) \\
		&\qquad\qquad\qquad +\sum_{k=1}^j E_{k,j}(s)E_{j,i}(r+1)E_{i,k}(-r-s-1) - \sum_{k=i+1}^N E_{i,j}(s)E_{k,i}(r+1)E_{j,k}(-r-s-1) \Bigg).
	\end{split}
\end{equation*}
The sums of the terms containing
\[
	E_{k,i}(a)E_{i,j}(b)E_{j,k}(c), \quad E_{k,j}(a)E_{j,i}(b)E_{i,k}(c)
\]
are
\begin{multline*}
		\sum_{r,s \geq 0} \Bigg( -\sum_{k=1}^i E_{k,i}(r+s+1)E_{i,j}(-r-1)E_{j,k}(-s) -\sum_{k=j+1}^N E_{k,i}(r+1)E_{i,j}(s)E_{j,k}(-r-s-1)\\
		+ \sum_{k=i+1}^N E_{k,j}(r+s+1)E_{j,k}(-r-s-1) \Bigg),
\end{multline*}
\begin{multline*}
		\sum_{r,s \geq 0} \Bigg( \sum_{k=j+1}^N E_{k,j}(r+s+1)E_{j,i}(-s)E_{i,k}(-r-1) +\sum_{k=1}^i E_{k,j}(s)E_{j,i}(r+1)E_{i,k}(-r-s-1)\\
		- \sum_{k=i+1}^N E_{k,j}(r+s+1)E_{j,k}(-r-s-1) \Bigg),
\end{multline*}
respectively.
Hence the third identity holds.

A similar computation shows the last identity.
\end{proof}
We show (\ref{eq:mainthm+-}).
Assume $i,j \neq 0$.
The proofs for the remaining cases are similar.
We have
\begin{equation*}
	\begin{split}
		&[\ev({x}_{i,1}^+), {x}_{j}^{-}] = (1+i \ve_2)\delta_{ij}h_i \\
		&\quad + \hbar \sum_{s \geq 0} \Big( \sum_{k=1}^i[E_{k,i+1}(s)E_{i,k}(-s), E_{j+1,j}] + \sum_{k=i+1}^N[E_{k,i+1}(s+1)E_{i,k}(-s-1), E_{j+1,j}] \Big).
	\end{split}
\end{equation*}
Since we have
\begin{equation*}
	\begin{split}
		&[E_{k,i+1}(s)E_{i,k}(-s), E_{j+1,j}] \\
		&= \Big(\delta_{i,j}E_{k,j}(s)-\delta_{j,k}E_{j+1,i+1}(s)\Big)E_{i,k}(-s) + E_{k,i+1}(s)\Big(\delta_{j+1,k}E_{i,j}(-s)-\delta_{i,j}E_{j+1,k}(-s)\Big),
	\end{split}
\end{equation*}
we see
\begin{equation*}
	\begin{split}
		&\sum_{k=1}^i[E_{k,i+1}(s)E_{i,k}(-s), E_{j+1,j}] \\
		&=\sum_{k=1}^i \delta_{i,j}E_{k,j}(s)E_{i,k}(-s) -\delta(j \leq i)E_{j+1,i+1}(s) E_{i,j}(-s) \\
		&\quad + \delta(j+1 \leq i)E_{j+1,i+1}(s) E_{i,j}(-s)- \sum_{k=1}^i \delta_{i,j}E_{k,i+1}(s)E_{j+1,k}(-s)\\
		&=\delta_{i,j} \Big( \sum_{k=1}^i \Big( E_{k,i}(s)E_{i,k}(-s) - E_{k,i+1}(s)E_{i+1,k}(-s) \Big) -E_{i+1,i+1}(s) E_{i,i}(-s) \Big)
	\end{split}
\end{equation*}
and
\begin{equation*}
	\begin{split}
		&\sum_{k=i+1}^N[E_{k,i+1}(s+1)E_{i,k}(-s-1), E_{j+1,j}] \\
		&=\sum_{k=i+1}^N \delta_{i,j}E_{k,j}(s+1)E_{i,k}(-s-1) -\delta(j \geq i+1)E_{j+1,i+1}(s+1) E_{i,j}(-s-1) \\
		&\quad + \delta(j \geq i)E_{j+1,i+1}(s+1) E_{i,j}(-s-1)- \sum_{k=i+1}^N \delta_{i,j}E_{k,i+1}(s+1)E_{j+1,k}(-s-1)\\
		&=\delta_{i,j} \Big( \sum_{k=i+1}^N \Big( E_{k,i}(s+1)E_{i,k}(-s-1) - E_{k,i+1}(s+1)E_{i+1,k}(-s-1) \Big) + E_{i+1,i+1}(s+1) E_{i,i}(-s-1) \Big).
	\end{split}
\end{equation*}
Therefore
\begin{equation*}
	\begin{split}
		&[\ev({x}_{i,1}^+), {x}_{j}^{-}] \\
		&= \delta_{i,j} \Bigg( (1+i \ve_2)h_i -\hbar E_{i+1,i+1} E_{i,i} + \hbar \sum_{s \geq 0} \Bigg( \sum_{k=1}^i \Big( E_{k,i}(s)E_{i,k}(-s) - E_{k,i+1}(s)E_{i+1,k}(-s) \Big) \\
		&\qquad\qquad\qquad\qquad\qquad\qquad + \sum_{k=i+1}^N \Big( E_{k,i}(s+1)E_{i,k}(-s-1) - E_{k,i+1}(s+1)E_{i+1,k}(-s-1) \Big) \Bigg) \Bigg) \\
		&= \delta_{i,j} \ev(h_{i,1}).
	\end{split}
\end{equation*}

We show (\ref{eq:mainthmHX}) for $+$.
Assume $i,j \neq 0$.
The right-hand side of (\ref{eq:mainthmHX}) is
\begin{equation*}
	\begin{split}
		&a_{ij} \Bigg( \Big( 1+i \ve_2 + (\delta_{i+1,j} - \delta_{i-1,j}) \dfrac{\hbar}{2} \Big) x_j^+ \\
		&\qquad\qquad + \hbar\sum_{s \geq 0} \Big( \sum_{k=1}^j E_{k,j+1}(s) E_{j,k}(-s) + \sum_{k=j+1}^N E_{k,j+1}(s+1) E_{j,k}(-s-1) \Big) \Bigg).
	\end{split}
\end{equation*}
A direct computation shows
\begin{equation*}
	\begin{split}
		&[\ev(\tilde{h}_{i,1}), x_{j}^{+}] = (1+ i \varepsilon_2) a_{ij} x_{j}^+ \\
		&\ - \dfrac{\hbar}{2} \Big( \delta_{i,j}(\{E_{i,i+1}, E_{i,i}\} - \{E_{i,i+1}, E_{i+1,i+1}\}) - \delta_{i,j+1}\{E_{i-1,i}, E_{i,i}\} + \delta_{i+1,j}\{E_{i+1,i+2}, E_{i+1,i+1}\} \Big)\\
		&\ \quad + \hbar \Big( \delta_{i,j}(E_{i,i}E_{i,i+1} - E_{i,i+1}E_{i+1,i+1}) - \delta_{i,j+1} E_{i,i}E_{i-1,i} + \delta_{i+1,j} E_{i+1,i+2}E_{i+1,i+1} \Big)\\
		&\ \qquad + \hbar \sum_{s \geq 0} \Bigg( 2\delta_{i,j} \Big( \sum_{k=1}^i E_{k,i+1}(s)E_{i,k}(-s) + \sum_{k=i+1}^N E_{k,i+1}(s+1)E_{i,k}(-s-1)  \Big)\\
		&\quad\qquad\qquad\qquad - \delta_{i,j+1} \Big( \sum_{k=1}^{i-1} E_{k,i}(s)E_{i-1,k}(-s) + \sum_{k=i}^N E_{k,i}(s+1)E_{i-1,k}(-s-1) \Big)\\
		&\qquad\qquad\qquad\qquad - \delta_{i+1,j} \Big( \sum_{k=1}^{i+1} E_{k,i+2}(s)E_{i+1,k}(-s) + \sum_{k=i+2}^N E_{k,i+2}(s+1)E_{i+1,k}(-s-1) \Big) \Bigg).
	\end{split}
\end{equation*}
Hence the desired identity follows from
\begin{equation*}
	\begin{split}
		&\dfrac{\hbar}{2} \Big(\!-\{E_{i,i+1}, E_{i,i}\} + \{E_{i,i+1}, E_{i+1,i+1}\} + 2E_{i,i}E_{i,i+1} - 2E_{i,i+1}E_{i+1,i+1}\Big) \\
		&=\dfrac{\hbar}{2}\Big([E_{i,i},E_{i,i+1}] + [E_{i+1,i+1},E_{i,i+1}]\Big) = 0
	\end{split}
\end{equation*}
for the case $i=j$,
\begin{equation*}
	\begin{split}
		&\dfrac{\hbar}{2} \Big(\{E_{i-1,i}, E_{i,i}\} - 2E_{i,i}E_{i-1,i}\Big) = \dfrac{\hbar}{2} [E_{i-1,i}, E_{i,i}] = \dfrac{\hbar}{2} x_{i-1}^+
	\end{split}
\end{equation*}
for the case $i=j+1$,
\begin{equation*}
	\begin{split}
		&\dfrac{\hbar}{2} \Big(\!-\{E_{i+1,i+2}, E_{i+1,i+1}\} + 2E_{i+1,i+2}E_{i+1,i+1}\Big) = \dfrac{\hbar}{2} [E_{i+1,i+2}, E_{i+1,i+1}] = -\dfrac{\hbar}{2} x_{i+1}^+
	\end{split}
\end{equation*}
for the case $i=j-1$.
We thus have proved the case $i,j \neq 0$.
The proofs for the remaining cases are similar, but we give computations for the cases $i=1$, $j=0$ and $i=0$, $j=1$ to clarify a role of the condition $\hbar c = N\ve_2$.
First assume $i=1$, $j=0$.
Then the right-hand side of (\ref{eq:mainthmHX}) is
\begin{equation*}
	\begin{split}
		&- \Bigg( \Big( 1+ \ve_2 + N \ve_2 - \dfrac{\hbar}{2} \Big) x_0^+ + \hbar\sum_{s \geq 0} \sum_{k=1}^N E_{k,1}(s+1) E_{N,k}(-s) \Bigg).
	\end{split}
\end{equation*}
The left-hand side is
\begin{equation*}
	\begin{split}
		&[\ev(\tilde{h}_{1,1}), x_{0}^{+}] \\
		&= -(1+ \varepsilon_2) x_{0}^+ + \dfrac{\hbar}{2} \{E_{N,1}(1), E_{1,1}\} -  \hbar c x_0^+ - \hbar E_{1,1}E_{N,1}(1) - \hbar \sum_{s \geq 0} \sum_{k=1}^N E_{k,1}(s+1)E_{N,k}(-s)\\
		&= - \Bigg( (1+ \ve_2)x_0^+  + \hbar c x_0^+ - \dfrac{\hbar}{2} \Big( \{E_{N,1}(1), E_{1,1}\}-2E_{1,1}E_{N,1}(1) \Big) + \hbar\sum_{s \geq 0} \sum_{k=1}^N E_{k,1}(s+1) E_{N,k}(-s) \Bigg).
	\end{split}
\end{equation*}
Hence the assertion holds.
Next assume $i = 0$, $j = 1$.
Then the right-hand side of (\ref{eq:mainthmHX}) is
\begin{equation*}
	\begin{split}
		&- \Bigg( \Big( 1+ \dfrac{\hbar}{2} \Big) x_1^+ + \hbar\sum_{s \geq 0} \Big( E_{1,2}(s) E_{1,1}(-s) + \sum_{k=2}^N E_{k,2}(s+1) E_{1,k}(-s-1) \Big) \Bigg).
	\end{split}
\end{equation*}
The left-hand side is
\begin{equation*}
	\begin{split}
		&[\ev(\tilde{h}_{0,1}), x_{1}^{+}] \\
		&= -(1+ N \varepsilon_2) x_{1}^+ - \dfrac{\hbar}{2} \Big( \{E_{1,2}, E_{1,1}\}-2E_{1,2}c \Big) - \hbar \sum_{s \geq 0} \sum_{k=1}^N E_{k,2}(s+1) E_{1,k}(-s-1)\\
		&= -\Bigg( (1+ N \varepsilon_2) x_1^+  - \hbar c x_1^+ + \dfrac{\hbar}{2} \Big( \{E_{1,2}, E_{1,1}\} - 2 E_{1,2}E_{1,1} \Big)\\
		&\qquad\qquad\qquad\qquad + \hbar \sum_{s \geq 0} \Big( E_{1,2}(s)E_{1,1}(-s) + \sum_{k=2}^{N} E_{k,2}(s+1)E_{1,k}(-s-1) \Big)\Bigg).
	\end{split}
\end{equation*}
Hence the assertion holds.

We show (\ref{eq:mainthmXX2}) for $+$.
First assume $i,j \neq 0$.
A direct computation shows
\begin{equation*}
	\begin{split}
		&[\ev(x_{i, 1}^{+}), x_{j}^{+}] = (1+ i \varepsilon_2) [x_{i}^{+},x_{j}^{+}] + \hbar \delta_{i,j} (x_i^+)^2\\
		&\qquad + \hbar \sum_{s \geq 0} \Bigg( \delta_{i+1,j} \Big( \sum_{k=1}^i   E_{k,i+2}(s) E_{i,k}(-s) + \sum_{k=i+1}^N E_{k,i+2}(s+1) E_{i,k}(-s-1) \Big) \\
		&\qquad\qquad\qquad - \delta_{i,j+1} \Big( \sum_{k=1}^i E_{k,i+1}(s)E_{i-1,k}(-s) + \sum_{k=i+1}^N E_{k,i+1}(s+1)E_{i-1,k}(-s-1) \Big)\Bigg).
	\end{split}
\end{equation*}
By swapping $i$ and $j$, and summing up, we obtain
\begin{equation*}
	\begin{split}
		&[\ev(x_{i, 1}^{+}), x_{j}^{+}] - [x_{i}^{+}, \ev(x_{j, 1}^{+})]= (i-j)\varepsilon_2 [x_{i}^{+},x_{j}^{+}] + 2 \hbar \delta_{i,j} (x_{i}^+)^2\\
		&\qquad + \hbar \Big( \delta_{i+1,j} (-x_{i+1}^+ x_{i}^+) + \delta_{i,j+1} (-x_i^+ x_{i-1}^+) \Big) \\
		&= \begin{cases}
			2 \hbar (x_{i}^+)^2 & j=i,\\
			-\ve_2 x_i^+ x_{i+1}^+ - \ve_1 x_{i+1}^+ x_{i}^+ & j=i+1,\\
			-\ve_1 x_i^+ x_{i-1}^+ - \ve_2 x_{i-1}^+ x_{i}^+ & j=i-1,\\
			0 & \text{otherwise}.
		\end{cases}
	\end{split}
\end{equation*}
These are nothing but the desired relations.

Next consider the other cases.
For the case $i=j=0$, we have
\begin{equation*}
	\begin{split}
		&[\ev(x_{0, 1}^{+}), x_{0}^{+}] = \hbar \sum_{s \geq 0} \sum_{k=1}^N [E_{k,1}(s+1) E_{N,k}(-s), E_{N,1}(1)]\\
		&= \hbar \sum_{s \geq 0} \Big(\! -E_{N,1}(s+2) E_{N,1}(-s) + E_{N,1}(s+1) E_{N,1}(-s+1) \Big) = \hbar (x_{0}^+)^2.
	\end{split}
\end{equation*}
For the case $i=0$, $j \neq 0$, we have
\begin{equation*}
	\begin{split}
		&[\ev(x_{0, 1}^{+}), x_{j}^{+}] - [x_{0}^{+}, \ev(x_{j, 1}^{+})]= (N-j)\varepsilon_2 [x_{0}^{+},x_{j}^{+}] - \hbar c \delta_{1,j} [x_{0}^{+},x_{1}^{+}]\\
		&\qquad + \hbar \Big( \delta_{1,j} (-x_{1}^+ x_{0}^+) + \delta_{N,j+1} (-x_0^+ x_{N-1}^+) \Big) \\
		&= \begin{cases}
			-\ve_2 x_0^+ x_{1}^+ - \ve_1 x_{1}^+ x_{0}^+ & j=1,\\
			-\ve_1 x_0^+ x_{N-1}^+ - \ve_2 x_{N-1}^+ x_{0}^+ & j=N-1,\\
			0 & \text{otherwise}.
		\end{cases}
	\end{split}
\end{equation*}
We need the condition $\hbar c = N \ve_2$ for the case $i=0$, $j=1$.
\end{proof}

\subsection{Cyclic automorphism}\label{subsection:cyclic}

We consider algebra automorphisms corresponding to the rotation of the Dynkin diagram.
It is easy to see that the assignment
\[
	x_i^{\pm} \mapsto x_{i-1}^{\pm}, \quad h_i \mapsto h_{i-1}, \quad c \mapsto c, \quad \bfid(s) \mapsto \bfid(s) + \delta_{s,0}c
\]
gives an algebra automorphism $\rho_U$ of $U(\hat{\gl}_N)$.
Guay introduced an analogous automorphism for the affine Yangian.
\begin{prop}[\cite{MR2199856}, Lemma~3.5]
The assignment
\[
	x_{i,r}^{\pm} \mapsto \sum_{s=0}^r \dbinom{r}{s} \ve_2^{r-s} x_{i-1,s}^{\pm}, \quad h_{i,r} \mapsto \sum_{s=0}^r \dbinom{r}{s} \ve_2^{r-s} h_{i-1,s}
\]
gives an algebra automorphism $\rho$ of $\affY$.
\end{prop}

\begin{lem}
We have $\rho_U(E_{ij}(s)) = E_{i-1,j-1}(s+\delta_{i,1}-\delta_{j,1}) + \delta_{s,0}\delta_{i,1}\delta_{j,1}c$.
\end{lem}
\begin{proof}
See \cite[Lemma~4.1]{kodera_braidgroup}.
\end{proof}

\begin{prop}
We have $\rho_{U} \circ \ev = \ev \circ \rho$.
\end{prop}

\begin{proof}
The identity obviously holds for $x_{i,0}^{\pm}$, $h_{i,0}$.
Hence it is enough to show that the identity also holds for $x_{i,1}^+$ since $\affY$ is generated by  $x_{i,0}^{\pm}$, $h_{i,0}$, $x_{i,1}^+$ ($i \in \bbZ/N\bbZ$).
We show $\rho_U(\ev(x_{i,1}^+)) = \ev(\rho(x_{i,1}^+))$ for $i \neq 0$.
The proof for the case $i = 0$ is similar.
The left-hand side is
\begin{equation*}
	\begin{split}
		&\rho_U(\ev(x_{i,1}^+)) = (1+ i \varepsilon_2) x_{i-1}^+ \\
		&\quad + \hbar \sum_{s \geq 0} \Big( \sum_{k=1}^i \rho_U(E_{k,i+1}(s) E_{i,k}(-s)) + \sum_{k=i+1}^N \rho_U(E_{k,i+1}(s+1) E_{i,k}(-s-1)) \Big).
	\end{split}
\end{equation*}
We have
\begin{equation*}
	\begin{split}
		&\sum_{k=1}^i \rho_U(E_{k,i+1}(s) E_{i,k}(-s)) =\sum_{k=1}^i  E_{k-1,i}(s+\delta_{k,1}) \Big( E_{i-1,k-1}(-s+\delta_{i,1}-\delta_{k,1}) + \delta_{s,0}\delta_{i,1}\delta_{k,1} c \Big) \\
		&= E_{N,i}(s+1) E_{i-1,N}(-s-1+\delta_{i,1}) + \delta_{s,0}\delta_{i,1} E_{N,1}(1)c + \sum_{k=2}^i E_{k-1,i}(s) E_{i-1,k-1}(-s+\delta_{i,1})
	\end{split}
\end{equation*}
and
\begin{equation*}
	\begin{split}
		&\sum_{k=i+1}^{N} \rho_U(E_{k,i+1}(s+1) E_{i,k}(-s-1)) =\sum_{k=i+1}^{N} E_{k-1,i}(s+1) E_{i-1,k-1}(-s-1+\delta_{i,1}).
	\end{split}
\end{equation*}
This verifies that $\rho_U(\ev(x_{i,1}^+))$ is equal to $\ev(\rho(x_{i,1}^+))=\ev(x_{i-1,1}^+)+\ve_2 x_{i-1}^+$ under the assumption $\hbar c = N \ve_2$.
\end{proof}

\subsection{Another version}

Guay's evaluation map $\ev$ has $U(\hat{\gl}_N)_{{\rm comp},-}$ as its target space and hence it matches with lowest weight modules of $\hat{\gl}_N$.
To deal with highest weight modules, we introduce an opposite evaluation map by using the anti-isomorphisms $\mu$ and $\mu_U$.
Moreover we make it one-parameter family of algebra homomorphisms by using $\tau_{\alpha}$.
\begin{dfn}
For each $\alpha \in \bbC$, define $\ev_\alpha^+ = \mu_U^{-1} \circ \ev \circ \tau_{\alpha-1} \circ \mu$.
Here $\ev$ and $\tau_{\alpha-1}$ are taken as those for $Y_{-\ve_2,-\ve_1}(\affsl)$.
\end{dfn}

\begin{thm}\label{thm:main}
Assume $\hbar c = -N \ve_1$ and let $\alpha$ be a complex number.
Then there exists an algebra homomorphism $\ev^+_{\alpha} \colon \affY \to U(\hat{\gl}_N)_{{\rm comp},+}$ uniquely determined by
\begin{gather*}
	\ev^+_{\alpha}(x_{i,0}^{+}) = x_{i}^{+}, \quad \ev^+_{\alpha}(x_{i,0}^{-}) = x_{i}^{-},\quad \ev^+_{\alpha}(h_{i,0}) = h_{i},
\end{gather*}
\begin{gather*}
	\ev^+_{\alpha}(x_{i,1}^{+}) = \begin{cases}
		(\alpha - N \varepsilon_1) x_{0}^{+} + \hbar \displaystyle\sum_{s \geq 0} \sum_{k=1}^N E_{N,k}(-s) E_{k,1}(s+1) \text{ if $i = 0$},\\
		(\alpha - i \varepsilon_1) x_{i}^{+} + \hbar \displaystyle\sum_{s \geq 0} \Big( \sum_{k=1}^i E_{i,k}(-s) E_{k,i+1}(s) + \sum_{k=i+1}^N E_{i,k}(-s-1) E_{k,i+1}(s+1) \Big)\\
		\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{ if $i \neq 0$},
	\end{cases}
\end{gather*}
\begin{gather*}
	\ev^+_{\alpha}(x_{i,1}^-) = \begin{cases}
		(\alpha - N \varepsilon_1) x_{0}^{-} + \hbar \displaystyle\sum_{s \geq 0} \sum_{k=1}^N E_{1,k}(-s-1) E_{k,N}(s) \text{ if $i = 0$},\\
		(\alpha - i \varepsilon_1) x_{i}^{-} + \hbar \displaystyle\sum_{s \geq 0} \Big( \sum_{k=1}^i E_{i+1,k}(-s) E_{k,i}(s) + \sum_{k=i+1}^N E_{i+1,k}(-s-1) E_{k,i}(s+1) \Big) \\
		\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{ if $i \neq 0$},
	\end{cases}
\end{gather*}
\begin{gather*}
	\ev^+_{\alpha}(h_{i,1}) = \begin{cases}
		(\alpha - N \varepsilon_1) h_{0} - \hbar E_{N,N} (E_{1,1}-c) \\
		\quad + \hbar \displaystyle\sum_{s \geq 0} \sum_{k=1}^{N} \Big( E_{N,k}(-s) E_{k,N}(s) - E_{1,k}(-s-1) E_{k,1}(s+1) \Big) \text{ if $i = 0$},\\
\\
		(\alpha - i \varepsilon_1) h_{i} - \hbar E_{i,i}E_{i+1,i+1} \\
		\quad + \hbar \displaystyle\sum_{s \geq 0} \Big( \sum_{k=1}^{i} E_{i,k}(-s) E_{k,i}(s) + \displaystyle\sum_{k=i+1}^{N} E_{i,k}(-s-1) E_{k,i}(s+1) \\
		\qquad\qquad\quad - \displaystyle\sum_{k=1}^{i} E_{i+1,k}(-s) E_{k,i+1}(s) - \displaystyle\sum_{k=i+1}^{N} E_{i+1,k}(-s-1) E_{k,i+1}(s+1) \Big) \\
		\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{ if $i \neq 0$}.
	\end{cases}
\end{gather*}
\end{thm}

\section{Evaluation modules}

Let $L(\Lambda)$ be the integrable irreducible highest weight module of $\hat{\gl}_N$ with highest weight $\Lambda$.
The dominant integral weight $\Lambda$ is determined by the data $\lambda_1, \ldots, \lambda_N \in \bbC$ and $K \in \bbZ_{\geq 1}$ satisfying $\lambda_i-\lambda_{i+1} \in \bbZ_{\geq 0}$ ($1 \leq i \leq N-1$) and $\lambda_N-\lambda_1+K \in \bbZ_{\geq 0}$.
The correspondence is given by
\[
	\langle E_{i,i}, \Lambda \rangle = \lambda_i \ (1 \leq i \leq N) \quad \text{ and } \quad \langle c, \Lambda \rangle = K.
\]
We denote a fixed highest weight vector of $L(\Lambda)$ by $v_{\Lambda}$.

Assume $K \hbar = -N \ve_1$.
We define a $\affY$-module $L(\Lambda,\alpha)$ by the pull-back of $L(\Lambda)$ via $\ev^+_{\alpha}$.
By abuse of notation, we regard $v_{\Lambda}$ as a vector of $L(\Lambda,\alpha)$.
The vector $v_{\Lambda}$ satisfies the condition of the highest weight vector for affine Yangian.
More precisely we have the following.
\begin{thm}\label{thm:highest_weight}
We have
\begin{gather*}
	x_{i,r}^+ v_{\Lambda} = 0,\\
	h_{i,r} v_{\Lambda} = a_i^r \langle h_{i}, \Lambda \rangle v_{\Lambda}
\end{gather*}
for all $i \in \bbZ/N\bbZ$ and $r \geq 0$, where
\begin{equation*}
	\begin{split}
		a_i &= \begin{cases}
			\alpha - N\ve_1 + \hbar \lambda_N & \text{if $i=0$},\\
			\alpha - i\ve_1 + \hbar \lambda_i & \text{otherwise},
		\end{cases}\\
		&= \begin{cases}
			\alpha + (\lambda_N+K)\hbar & \text{if $i=0$},\\
			\alpha + \left(\lambda_i+\dfrac{i}{N}K \right)\hbar & \text{otherwise}.
	\end{cases}
	\end{split}
\end{equation*}
\end{thm}
The goal of the remaining part is to prove this theorem.

\begin{rem}\label{rem:thm}
In \cite{kodera_braidgroup}, the author proves that the image of Guay's evaluation map $\ev$ in $U(\hat{\gl}_N)_{{\rm comp},-}$ contains $\hat{\gl}_N$ under the assumption $\ve_2 \neq 0$.
This result implies that the image of $\ev^+_{\alpha}$ in $U(\hat{\gl}_N)_{{\rm comp},+}$ contains $\hat{\gl}_N$ under the assumption $\ve_1 \neq 0$.
Hence the $\affY$-module $L(\Lambda,\alpha)$ is irreducible when $\ve_1 \neq 0$.
We do not use this fact in the proof given below.
\end{rem}

We provide a general lemma.
\begin{lem}\label{lem:general_lemma}
Let $v$ be a nonzero element of a $\affY$-module satisfying
\begin{equation}
	x_{i,0}^+ v = 0, \quad h_{i,0} v = P_{i,0} v \label{eq:assumption1}
\end{equation}
for some $P_{i,0} \in \bbC$.
Further assume that $v$ satisfies
\begin{equation}
	h_{i,1} v = P_{i,1} v, \label{eq:assumption2}
\end{equation}
\begin{equation}
	x_{i,1}^- v = Q_{i} x_{i,0}^- v \label{eq:assumption3}
\end{equation}
for some $P_{i,1}, Q_{i} \in \bbC$.
Then $v$ satisfies
\[
	x_{i,r}^+ v = 0,\quad h_{i,r} v = P_{i,r} v
\]
for all $r \geq 0$, where $P_{i,r}$ $(r \geq 2)$ is given by
\begin{equation}
	P_{i,r} = Q_{i}^{r-1} P_{i,1}. \label{eq:eigen}
\end{equation}
\end{lem}

\begin{proof}
We prove $x_{i,r}^+ v = 0$ by induction on $r$.
The assertion for $r=0$ holds by the assumption~(\ref{eq:assumption1}).
Assume it for $r$.
Then we have
\begin{equation*}
	\begin{split}
		x_{i,r+1}^+ v = \dfrac{1}{2} [\tilde{h}_{i,1}, x_{i,r}^+] v = -\dfrac{1}{2} x_{i,r}^+ \tilde{h}_{i,1} v = -\dfrac{1}{2} x_{i,r}^+ \left( h_{i,1}-\dfrac{\hbar}{2}h_{i,0}^2 \right) v
	\end{split}
\end{equation*}
by the induction assumption, and it is equal to $0$ by the assumptions~(\ref{eq:assumption1}), (\ref{eq:assumption2}) and again by the induction assumption.

We prove that $v$ is an eigenvector of $h_{i,r}$ with the eigenvalue (\ref{eq:eigen}) by induction on $r$.
The assertion for $r=1$ holds by the assumption (\ref{eq:assumption2}).
Assume it for $r$.
We have
\begin{equation*}
	\begin{split}
		h_{i,r+1} v &= [x_{i,r+1}^+,x_{i,0}^-] v = x_{i,r+1}^+ x_{i,0}^- v = \dfrac{1}{2}[\tilde{h}_{i,1},x_{i,r}^+] x_{i,0}^- v\\
		&= \dfrac{1}{2} \left( \tilde{h}_{i,1} x_{i,r}^+ x_{i,0}^- v - x_{i,r}^+ \tilde{h}_{i,1} x_{i,0}^- v \right).
	\end{split}
\end{equation*}
Here we use $x_{i,r+1}^+ v = 0$ in the second equality.	Then we have
\begin{equation*}
	\begin{split}
		\tilde{h}_{i,1} x_{i,r}^+ x_{i,0}^- v &= \tilde{h}_{i,1} ([x_{i,r}^+, x_{i,0}^-] + x_{i,0}^- x_{i,r}^+ )v \\
		&= \tilde{h}_{i,1} h_{i,r} v \quad \text{by $x_{i,r}^+ v=0$} \\
		&= \left(P_{i,1}-\dfrac{\hbar}{2}P_{i,0}^2 \right) P_{i,r} v \quad \text{by (\ref{eq:assumption1}), (\ref{eq:assumption2}), the induction assumption,}
	\end{split}
\end{equation*}
and
\begin{equation*}
	\begin{split}
		x_{i,r}^+ \tilde{h}_{i,1} x_{i,0}^- v &= x_{i,r}^+ ([\tilde{h}_{i,1}, x_{i,0}^-] + x_{i,0}^- \tilde{h}_{i,1} )v \\
		&= -2 x_{i,r}^+ x_{i,1}^- v + \left(P_{i,1}- \dfrac{\hbar}{2} P_{i,0}^2\right)x_{i,r}^+ x_{i,0}^- v \quad \text{by (\ref{eq:assumption1}), (\ref{eq:assumption2})}\\
		&= \left( -2Q_{i} + P_{i,1}- \dfrac{\hbar}{2} P_{i,0}^2\right)x_{i,r}^+ x_{i,0}^- v \quad \text{by (\ref{eq:assumption3})}\\
		&= \left( -2Q_{i} + P_{i,1}- \dfrac{\hbar}{2} P_{i,0}^2 \right)h_{i,r} v \quad \text{by $x_{i,r}^+ v=0$}\\
		&= \left( -2Q_{i} + P_{i,1}- \dfrac{\hbar}{2} P_{i,0}^2 \right) P_{i,r} v \quad \text{by the induction assumption.}
	\end{split}
\end{equation*}
Hence we conclude $h_{i,r+1}v = Q_{i} P_{i,r} v$, which completes the proof.
\end{proof}

The vector $v_{\Lambda}$ satisfies the condition (\ref{eq:assumption1}).
In order to apply Lemma~\ref{lem:general_lemma} to our situation, we compute $h_{i,1} v_{\Lambda}$ and $x_{i,1}^- v_{\Lambda}$.
\begin{lem}\label{lem:r=1}
We have $h_{i,1} v_{\Lambda} = a_i \langle h_{i}, \Lambda \rangle v_{\Lambda}$ for all $i \in \bbZ/N\bbZ$.
\end{lem}
\begin{proof}
We give a proof for $i\neq 0$.
The case $i=0$ is similar.
Recall the formula for $\ev^+_{\alpha}(h_{i,1})$ and note that
 all the terms in $\ev^+_{\alpha}(h_{i,1})$ annihilate $v_{\Lambda}$ except for those concerning
\[
	h_i v_{\Lambda} = \langle h_{i}, \Lambda \rangle v_{\Lambda},
\]
\[
	E_{i,i}E_{i+1,i+1} v_{\Lambda} = \lambda_i \lambda_{i+1} v_{\Lambda},
\]
\[
	E_{i,i}^2 v_{\Lambda} = \lambda_i^2 v_{\Lambda}.
\]
Then we see that $v_{\Lambda}$ is an eigenvector of $h_{i,1}$ with the eigenvalue
\begin{equation*}
	(\alpha - i\ve_1)\langle h_{i}, \Lambda \rangle -\hbar \lambda_i \lambda_{i+1} + \hbar \lambda_i^2 = (\alpha - i\ve_1 + \hbar \lambda_i)\langle h_{i}, \Lambda \rangle.
\end{equation*}
\end{proof}

\begin{lem}\label{lem:wt-2}
We have $x_{i,1}^- v_{\Lambda} = a_i x_i^- v_{\Lambda}$ for all $i \in \bbZ/N\bbZ$.
\end{lem}
\begin{proof}
We give a proof for $i\neq 0$.
The case $i=0$ is similar.
Recall the formula for $\ev^+_{\alpha}(x_{i,1}^-)$.
Then we have
\begin{equation*}
	\begin{split}
		x_{i,1}^- v_{\Lambda} = (\alpha - i\ve_1) x_i^- v_{\Lambda} + \hbar E_{i+1,i}E_{i,i} v_{\Lambda} = (\alpha - i\ve_1 + \hbar \lambda_i) x_i^- v_{\Lambda}.
	\end{split}
\end{equation*}
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:highest_weight}]

By Lemma~\ref{lem:general_lemma}, \ref{lem:r=1}, \ref{lem:wt-2}, we see that $x_{i,r}^+ v_{\Lambda}=0$ and $v_{\Lambda}$ is an eigenvector of $h_{i,r}$ with the eigenvalue
\[
	a_i^{r-1} \times a_i \langle h_{i}, \Lambda \rangle = a_i^r \langle h_{i}, \Lambda \rangle.
\]
\end{proof}

Let us consider the generating series
\[
	1 + \hbar \sum_{r=0}^{\infty} a_i^{r} \langle h_{i}, \Lambda \rangle u^{-r-1} \in \bbC[[u^{-1}]]
\]
of the eigenvalues of $h_{i,r}$ on $v_{\Lambda}$.
It is equal to
\[
	\dfrac{u-a_i+\langle h_{i}, \Lambda \rangle \hbar}{u-a_i} = \dfrac{\pi_i(u+\hbar)}{\pi_i(u)}
\]
where we put $\pi_i(u) = (u-a_i)(u-a_i+\hbar)\cdots(u-a_i+(\langle h_{i}, \Lambda \rangle-1)\hbar)$.
The collection $(\pi_i(u))$ is an analog of the Drinfeld polynomials of finite-dimensional irreducible modules for the Yangian associated with a simple Lie algebra.

In \cite{kodera_prims, kodera_TG}, the author constructed the level-one Fock representation $F$ of $\affY$.
The polynomials ($\pi_i(u)$) corresponding to its highest weight is
\[
	\pi_i(u)=\begin{cases}
	u & \text{if $i=0$},\\
	1 & \text{otherwise}
	\end{cases}
\]
by \cite[Theorem~5.7]{kodera_prims} (the highest weight vector is denoted by $b_{\varnothing}$).
We see that the level-one Fock representation at $\hbar = -N\ve_1$ with $\ve_1 \neq 0$ is an example of evaluation modules.
Indeed, take $\Lambda$ as the level-one dominant integral weight corresponding to $\lambda_1 = \cdots = \lambda_N=0$ and put $\alpha=-\hbar$.
Then the highest weights of $F$ and $L(\Lambda, -\hbar)$ are the same.
As we mentioned in Remark~\ref{rem:thm}, $L(\Lambda, -\hbar)$ is irreducible.
Since $F$ and $L(\Lambda)$ are isomorphic as $\affsl$-modules, $F$ and $L(\Lambda, -\hbar)$ are isomorphic as $\affY$-modules.

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\providecommand{\MRhref}[2]{
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
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TITLE: What is first Chern class and how to calculate it?
QUESTION [1 upvotes]: Apparently Calabi-Yau manifolds have vanishing first Chern class. Is there some simple explanation for what first Chern class of a manifold means? How to calculate it for a given metric (in physicist notation)?

REPLY [0 votes]: There are plent of books and review articles about Chern classes for physicists, but it would help to know more about what you already know before I can make recommendations. The fact that you mention  an advanced topic like Calabi-Yau manifolds while not knowing  about much simpler stuff like Chern classes makes me wonder what it is you are studying, and at what stage you are at. Can you give a hint? -- then I can give much better advice.
One thing is useful to know that is that Chern classes characterise line bundles, and for most types of manifolds a line bundle(a U(1) gauge field)  is something extra. Calabi-Yau manifolds, however, are Kahler manifolds and they have a built-in line bundle. Do you know what a Kahler manifold is?


TITLE: Doubt about an improper multiple integral
QUESTION [1 upvotes]: Study the convergence of the following integral $$\iint_A\frac{dxdy}{x+y}$$
Where $A=[0,1]^2$. If we consider  $C=\{(x,y)\in \mathbb R : x^2+y^2\le1,\ x>0, y>0 \}$, we can say that $$\iint_A\frac{dxdy}{x+y}=\iint_{A-C}\frac{dxdy}{x+y}+\iint_C\frac{dxdy}{x+y}$$
Now the integral on $A-C$ is a proper integral, while the integral on $C$ is improper in $(x,y)=(0,0)$. Applying polar coordinates:
$$\iint_C \frac{dxdy}{x+y}=\lim_{\varepsilon \to 0^+}\int_\varepsilon^1d\rho\int_0^{\frac \pi2}\frac{d\theta}{\cos \theta + \sin \theta}=\lim_{\varepsilon\to0^+}(1-\varepsilon)\int_0^{\frac \pi2}\frac{d\theta}{\cos \theta + \sin \theta}$$
Now we can say that $\cos \theta + \sin \theta$ is a continuous function and $C$ is compact, so for the Weierstrass theorem $\exists m>0 : \cos \theta + \sin \theta \ge m, \ \forall \theta \in[0,\frac \pi2]$, and he concludes that this last integral is convergent. I've doubts about:
1) Why do we need to say all the stuff about $m$ in the end? It's because we're doing a limit in two dimensions so we have to prove that it is uniform in $\theta$ so it exists? Or is for another reason?
2) After this affirmation about $m$ he just applies squeeze theorem like this $$\lim_{\varepsilon\to0^+}(1-\varepsilon)\int_0^{\frac \pi2}\frac{d\theta}{\cos \theta + \sin \theta}\le\lim_{\varepsilon\to0^+}(1-\varepsilon)\int_0^{\frac \pi2}\frac{d\theta}{m}$$
And since the last integral is convergent he concludes that the other one is convergent too, or is he using something else?
Thanks for your time.

REPLY [2 votes]: (2) is the reason for (1).
We need to affirm that (1) holds so that we can say (2) does.

$$\exists m{>}0\ ~\forall \theta\in[0;\pi/2] : (m\leq \cos \theta+\sin\theta)\tag 1$$
Which is true since the trig. sum is strictly positive over that interval.

$$0~\leq ~\lim_{\epsilon\to 0^+}(1-\epsilon)\int_0^{\pi/2}\dfrac{\mathrm d\theta }{\cos\theta+\sin\theta}~\leq~ \lim_{\epsilon\to 0^+}(1-\epsilon)\int_0^{\pi/2}\dfrac{\mathrm d\theta }{m}\tag 2$$
Which means the integral we seek must be less than a finite value, whatever it is.


TITLE: What causes a car wheel to skid under heavy braking?
QUESTION [0 upvotes]: Say you are driving down the road and and jam on the brakes in a car with no ABS, at what point does the wheel stop spinning and stop causing a skid.
I suspect it is when the friction between the brake disk and brake pad become greater than the friction between the tyre and the road?
Just want to make sure I am thinking about this in the correct way.
Thanks

REPLY [1 votes]: It is the static friction between the bottom of the tire and the road surface that keeps the tire from slipping or skidding (kinetic friction). Even though the tire is rotating, the bottom of a non slipping tire is in static contact with the road. When applied force is greater than static friction the wheel will slip. This can be caused by hard braking (skidding), extreme applied drive train power (peeling out), extreme cornering (sliding sideways), or a combination of these forces. Rain, ice, oil, or other properties can reduce static friction co-efficient  between the tire and the road, making slipping easier.

REPLY [0 votes]: . . . . at what point does the wheel stop spinning and stop causing a skid?

If the wheels are not rotating and the car is moving then the car will be in a skid.
The condition that there is no relative movement at the point of contact between the road and the tyre is that $v=r \omega$ where $v$ is the speed of the wheel axles, $r$ is the radius of the wheel and $\omega$ is angular speed of the wheel.
When the brakes are applied the effect is that frictional forces due to the brake pads reduce the speed of rotation of the wheels which in turn means that there are frictional forces acting on the tyres due to the road in the opposite direction to the motion of the car.
If the condition $v=r \omega$ is maintained, ie there is no relative movement at the point of contact, then the frictional forces between the tyre and the road are static.
Applying the brakes harder, ie slowing the car down faster, requires the static frictional forces to be larger.
However there is a limit to the magnitude of the static frictional forces related to the coefficient of static friction $\mu_{\rm static}$.
If the maximum value of the static frictional force is insufficient to produce the required rate of reduction of speed then there will be relative movement between the tyre and the road at the point of contact, ie $v>r\omega$, and a skid occurs.
The important parameter is now the coefficient of kinetic friction, $\mu_{\rm kinetic}$, which is smaller than $\mu_{\rm static}$ so at the onset of a skid there is an almost instantaneous decrease in the frictional force trying to slow the car down which can lead to a loss of control of the car.


TITLE: Do circles divide the plane into more regions than lines?
QUESTION [55 upvotes]: In this post it is mentioned that $n$ straight lines can divide the plane into a maximum number of $(n^{2}+n+2)/2$ different regions.
What happens if we use circles instead of lines? That is, what is the maximum number of regions into which n circles can divide the plane?
After some exploration it seems to me that in order to get maximum division the circles must intersect pairwise, with no two of them tangent, none of them being inside another and no three of them concurrent (That is no three intersecting at a point).
The answer seems to me to be affirmative, as the number I obtain is $n^{2}-n+2$ different regions. Is that correct?

REPLY [8 votes]: A qualitative approach:
Consider a group of $n$ lines that divide the plane into maximally many regions. The intersections between the lines all fall within a bounded area of the plane. Now replaces each of the lines with a huge circle, so large that within the bounded area-of-intersections it deviates from a straight line by much less than the distance between two points of intersection.
Then you still have all the regions you had with lines. However, if there are at least three circles, you also get some new regions that are far away from the central bounded area.


TITLE: residue theorem with logarithmic function
QUESTION [4 upvotes]: I have problem integrating function with logarithm.
Problems seems always to be branch cut of $\log$, but here it is different I think.
I have task to integrate
$$\oint_{|z| = 1} \! dz \log\left(\frac{z - a}{z - b}\right)$$
given $|a| < 1$ and $|b| < 1$
First I think to check branch points of logarithm.
I write $$\log\left(\frac{z - a}{z - b}\right) = \log(z - a) - \log(z - b)$$
Going around $a$ in small circle I pick up term $2\pi i$, going around $b$ I pick up term $-2\pi i$, so going around both I pick up nothing.
So I make branch cut $a$ to $b$ and contour does not intersect.
But now I get stuck! I try to find residue of integral at $a$ and $b$ but I cannot get series of at $a,b$, because logarithm always remains inside.
How I can evaluate such integrals?

REPLY [1 votes]: METHOD 1:
For the branch points at $z=a$ and $z=b$, we choose branch cuts as straight line contours, that begin at the branch points, intersect the unit circle at $e^{i\phi_a}$ and $e^{i\phi_b}$, then extend to the point at infinity.  Note that these branch cuts are not unique.
Now, let $z=e^{it}$.  Then, we have
$$\begin{align}
\oint_{|z|=1} \log(z-a)\,dz &= \int_0^{\phi_a^-}\log(e^{it})ie^{it}\,dt+ \int_{\phi_a^+}^{2\pi}\log(e^{it})ie^{it}\,dt\\\\
&=\left.\left((e^{it}-a)\left(\log(e^{it}-a)-1\right)\right)\right|_{0}^{\phi_a^-}+\left.\left((e^{it}-a)\left(\log(e^{it}-a)-1\right)\right)\right|_{\phi_a^+}^{2\pi}\\\\
&=2\pi i (e^{i\phi_a}-a)
\end{align}$$
Similarly, we have
$$\oint_{|z|=1} \log(z-b)\,dz =2\pi i (e^{i\phi_b}-b)$$
Putting it together, we have
$$\oint_{|z|=1} \log\left(\frac{z-a}{z-b}\right)\,dz =2\pi i (e^{i\phi_a}-e^{i\phi_b}+(b-a))$$
If we choose $\phi_a=\phi_b$, then the integral of interest becomes
$$\oint_{|z|=1} \log\left(\frac{z-a}{z-b}\right)\,dz =2\pi i (b-a) \tag1$$

METHOD 2:
Here, we note that if the branch cut is chosen to adjoin the branch points, then the integrand is analytic outside $|z|=1$.  Therefore, we can evaluate the integral of interest using the Residue Theorem with the Residue at Infinity.  To that end, we have
$$\begin{align}
\oint_{|z|=1} \log\left(\frac{z-a}{z-b}\right)\,dz &=-2\pi i \text{Res}\left(-\frac{1}{z^2}\log\left(\frac{z^{-1}-a}{z^{-1}-b}\right),z=0\right)\\\\
&=2\pi i\lim_{z\to 0} \left(\frac{1}{z}\log\left(\frac{1-az}{1-bz}\right)\right)\\\\
&=2\pi i (b-a)
\end{align}$$
recovering the result in $(1)$!


TITLE: A divergent series i can't manage to sum
QUESTION [2 upvotes]: I know that this series is not summable in the usual sense so I'm "playing the game" of divergent series summation. The series is:
$$\sum_{n=0}^{+\infty}-\frac 12(1+3(-1)^{n+1})=1-2+1-2+1-2+\dots$$
I tried Abel summation method which failed this cause:
$$\sum_{n=0}^{+\infty}-\frac 12(1+3(-1)^{n+1})x^n=1-2x+x^2-2x^3+\dots=\frac {1-2x}{1-x^2}$$
Has singularities at $x=\pm 1$
Then i tried Borel summation but again I get to:
$$\sum_{n=0}^{+\infty}-\frac 12(1+3(-1)^{n+1})x^n=\frac 12(\frac 1{x-1}-\frac 3{x+1})$$
For $x\in ]-1;+\infty[-\{1\}$ that gives the same problem, and I don't think this is the case for Ramanujan summation.
I started to try Cesaro summation and I got to the following sequence of partial sums:
$$1;-1;0;-2;-1;-3;-2;-4;\dots$$
And the sequence of Cesaro means seems to diverge:
$$1;0;0;-\frac 12;-\frac 35;-1;-\frac 87;-\frac 32;\dots$$
I really don't know what other method should I try.

REPLY [1 votes]: You might try Zeta function regularization.
EDIT: Warning: depending on how you do this, you can get different results.  One way is this:
$$\sum_{n=1}^\infty -\frac{1}{2} (1 + 3 (-1)^n) n^{-s} =\zeta(s) - \dfrac{3}{2^s} \zeta(s)$$
and taking $s=0$ gives us the value $1$.


TITLE: $f,g \in R(T)$ such that $\hat{f} \cdot n^{2/3} = \hat{g}$ prove that $f$'s Fourier series converges absolutely.
QUESTION [1 upvotes]: Can someone help me by checking my solution. Is there a shorter More elegant solution ?(i'm almost sure you can some how express  $f$'s Fourier series using $|\hat{g}|^2$ + constant, i saw someone do it in few easy manipulations but was never able to reproduce it myself)
$f,g \in R(T)$ (integrable periodic functions with period $2\pi$) such that $\hat{f} \cdot n^{2/3} = \hat{g}$ prove that $f$'s Fourier series converges absolutely.
$$(S_{N}f)(t) = \sum\limits_{n=-N}^N \hat{g}(n) \cdot \frac{e^{int}}{n^{2/3}}$$
$$|\hat{g'(n)}|^2 = n^2|\hat{g(n)}|^2$$
$$\sum n^2|\hat{g}(n)|^2 < \infty$$
$$|\hat{g(n)}| = |\hat{g(n)}| \cdot n \cdot \frac{1}{n} \le \frac{1}{2} \cdot(|\hat{g(n)}|^2n^2 + \frac{1}{n^2})\cdot n$$
the last Inequality is just means Inequality. Now finally
$$\sum|\hat{g(n)}| \le \frac{1}{2}\cdot \sum n^2|\hat{g(n)}|^2 + |\hat{g(0)}| + \frac{\pi^2}{6} < \infty$$
so
$$\sum\limits_{n=-\infty}^{\infty}|\hat{g}(n)| < \infty$$
moreover $\lim\limits_{n \rightarrow \infty}|\frac{e^{int}}{n^{2/3}}| = 0$ now we can use dirichlet's test for convergence of functions series to finish the proof.
Is it correct ?
I'm almost sure that you can solve this by expressing $f$'s Fourier series with $(\hat{g}(n))^2$  and then using parseval's theorem to prove that the series converges .

REPLY [2 votes]: It is enough to show that $\sum|\hat f(n)|<\infty$.
$$
\sum|\hat f(n)|=\sum n^{-2/3}|\hat g(n)|\le\Bigl(\sum n^{-4/3}\Bigr)^{1/2}\Bigl(\sum|\hat g(n)|^2\Bigr)^{1/2}<\infty
$$
since $4/3>1$ and $g$ is Riemann-integrable, and hence in $L^2$.


text:
\begin{document}

\title{Constructions of Optimal and Almost Optimal Locally Repairable Codes}

\author{
\authorblockN{Toni Ernvall}
\authorblockA{Turku Centre for Computer Science, Turku, Finland\\ \&  Department of Mathematics and Statistics\\ FI-20014 University of Turku, Finland\\ (e-mail: tmernv@utu.fi)}\and
\authorblockN{Thomas Westerb{\"a}ck and Camilla Hollanti }
\authorblockA{Department of Mathematics and Systems Analysis\\ Aalto University, P.O. Box 11100\\ FI-00076 Aalto, Finland \\ (e-mails: \{firstname.lastname@aalto.fi\})}
}

\maketitle

\newtheorem{definition}{Definition}[section]
\newtheorem{thm}{Theorem}[section]
\newtheorem{proposition}[thm]{Proposition}
\newtheorem{lemma}[thm]{Lemma}
\newtheorem{corollary}[thm]{Corollary}
\newtheorem{exam}{Example}[section]
\newtheorem{conj}{Conjecture}
\newtheorem{remark}{Remark}[section]

\newcommand{\La}{\mathbf{L}}
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\newcommand{\R}{{\mathbf R}}
\newcommand{\C}{{\mathbf C}}
\newcommand{\D}{{\mathcal D}}
\newcommand{\F}{{\mathbf F}}
\newcommand{\HH}{{\mathbf H}}
\newcommand{\OO}{{\mathcal O}}
\newcommand{\G}{{\mathcal G}}
\newcommand{\A}{{\mathcal A}}
\newcommand{\B}{{\mathcal B}}
\newcommand{\I}{{\mathcal I}}
\newcommand{\E}{{\mathcal E}}
\newcommand{\PP}{{\mathcal P}}
\newcommand{\Q}{{\mathbf Q}}
\newcommand{\M}{{\mathcal M}}
\newcommand{\separ}{\,\vert\,}
\newcommand{\abs}[1]{\vert #1 \vert}

\begin{abstract}
Constructions of optimal locally repairable codes (LRCs) in the case of $(r+1) \nmid n$ and over small finite fields were stated as open problems for LRCs in  [I. Tamo \emph{et al.}, ``Optimal locally repairable codes and connections to matroid theory'', \emph{2013 IEEE ISIT}].  In this paper, these problems are studied by constructing almost optimal linear LRCs, which are proven to be optimal for certain parameters, including cases for which $(r+1) \nmid n$. More precisely, linear codes for given length, dimension, and all-symbol locality are constructed with almost optimal minimum distance. `Almost optimal' refers to the fact that their minimum distance differs by at most one from the optimal value given by a known bound for LRCs. In addition to these linear LRCs, optimal LRCs which do not require a large field are constructed for certain classes of parameters.
\end{abstract}

\section{Introduction}
\subsection{Locally Repairable Codes}
In the literature, three kinds of repair cost metrics are studied: \emph{repair bandwidth} \cite{dimakis}, \emph{disk-I/O} \cite{diskIO}, and \emph{repair locality} \cite{Gopalan,Oggier,Simple}. In this paper the repair locality is the subject of interest.

 Given a finite field $\mathbb{F}_q$ with $q$ elements and an injective function $f: \mathbb{F}_q^k \rightarrow \mathbb{F}_q^n$, let $C$ denote the image of $f$. We say that $C$ is a \emph{locally repairable code (LRC)} and has  \emph{all-symbol locality} with parameters $(n,k,r,d)$, if the code $C$ has minimum (Hamming) distance $d$ and all the $n$ symbols of the code have repair locality $r$. The $j$th symbol has repair locality $s$ if there exists a set
$$
\{i_1,\ldots,i_s\} \subseteq \{1,\ldots,n\} \setminus \{j\}
$$
and a function $f_j$ such that
$$
f_j((y_{i_1}, \ldots, y_{i_s})) = y_j \hbox{ for all } \boldsymbol{y} \in C.
$$
LRCs are defined when $1 \leq r \leq k$. By a linear LRC we mean a linear code of length $n$ and dimension $k$.

In \cite{LRCpapailiopoulos}, Papailiopoulos \emph{et al.} establish an information theoretic bound for both linear and nonlinear codes. With $\epsilon = 0$ in \cite[Thm. 1]{LRCpapailiopoulos} we have the following bound for a locally repairable code $C$ with parameters $(n,k,r,d)$:
\begin{equation}
\label{upperbound}
d \leq n - k - \left \lceil \frac{k}{r} \right \rceil + 2
\end{equation}
A locally repairable code that meets this bound is called \emph{optimal}.

\subsection{Related Work}
As mentioned above, in the all-symbol locality case the information theoretic trade-off between locality and code distance for any (linear or nonlinear) code was derived in \cite{LRCpapailiopoulos}. Furthermore, constructions of optimal LRCs for the case when $(r+1) \nmid n$ and over small finite fields when $k$ is large were stated as open problems for LRCs in \cite{LRCmatroid}. In \cite{LRCmatroid} it was proved that there exists an optimal LRC for parameters $(n,k,r)$ over a field $\mathbb{F}_q$ if $r+1$ divides $n$ and $q=p^{k+1}$ with $p$  large enough. In \cite{SongOptimal} and \cite{TamoBarg} the existence of optimal LRCs was proved for several parameters $(n,k,r)$. Good codes with the weaker assumption of information symbol locality are designed in \cite{Pyramid}. In \cite{Gopalan} it was shown that there exist parameters $(n,k,r)$ for linear LRCs for which the bound of Eq. \eqref{upperbound} is not achievable.

\subsection{Contributions and Organization}
In this paper, we try to build good codes with all-symbol locality, when given parameters $n$, $k$, and $r$. As a measure for the goodness of a code we use its minimum distance $d$. Also, we prefer codes with simple structure, and the property that the construction does not require large field size. Moreover, we give some constructions of optimal LRCs, including cases for which $(r+1) \nmid n$, as well as constructions over small fields. Although codes in the case $(r+1) \nmid n$ are already constructed in \cite{SongOptimal} and \cite{TamoBarg}, the benefits of our construction are that it uses only some elementary linear algebra and it is very simple.

Section \ref{Sec:minDistance} studies the largest achievable minimum distance of the linear locally repairable codes. We show that with a field size large enough we have linear codes with minimum distance at least $d_{\text{opt}}(n,k,r)-1$ for every feasible triplet of parameters $(n,k,r)$. In  Subsection \ref{Subsec:Construction}, we give a construction of such an almost optimal linear locally repairable code. In Subsection \ref{Subsec:Analysis}, we analyze the minimum distance of our construction and derive a lower bound for the largest achievable minimum distance of the linear locally repairable code. Moreover, we prove that our construction results in optimal LRCs (including cases of $(r+1) \nmid n$) for specific parameter values.

In Section \ref{Subsec:Construction-Optimal} we give some constructions of optimal LRCs for certain classes of parameters which do not require a large field. Namely, for certain values of $(r,d)$, we give constructions of optimal $(n,k,r,d)$-LRCs for which the size of the field does not depend on the size of $k$ and $n$.

\section{Constructing Almost Optimal Codes}\label{Sec:minDistance}

\subsection{Construction}\label{Subsec:Construction}
In this subsection we will give a construction for linear locally repairable codes with all-symbol locality over a field $\mathbb{F}_q$ with $q>2\binom{n}{k-1}$, given parameters $(n,k,r)$ such that $n-\left\lceil\frac{n}{r+1}\right\rceil \geq k$. We also assume that $k<n$ and $n \not\equiv 1 \mod r+1$. Write $n=a(r+1)+b$, where $0 \leq b < r+1$.
We will construct a generator matrix for a linear code under the above assumptions. The minimum distance of the constructed code will be studied in Subsection \ref{Subsec:Analysis}.

Next we will build $A=\left\lceil\frac{n}{r+1}\right\rceil$ sets $S_1, S_2, \dots, S_A$ such that each of them consists of $r+1$ vectors of $\mathbb{F}_q^k$, except for $S_A$ that shall consist of $n-(A-1)(r+1)$ vectors of $\mathbb{F}_q^k$.

First, choose any $r$ linearly independent vectors $\mathbf{g}_{1,1},\dots,\mathbf{g}_{1,r}$. Let $\mathbf{s}_{1,r+1}$ be $\sum_{l=1}^{r}\mathbf{g}_{1,l}$. These $r+1$ vectors form the set $S_1$. This set has the property that any $r$ vectors from this set are linearly independent.

Let $1< i \leq A$. Assume that we have $i-1$ sets $S_1, S_2, \dots, S_{i-1}$ such that when taken at most $k$ vectors from these sets, at most $r$ vectors from each set, these vectors are linearly independent. Next we will show inductively that this is possible by constructing the set $S_i$ with the same property.

Let $\mathbf{g}_{i,1}$ be any vector such that when taken at most $k-1$ vectors from the already built sets, with at most $r$ vectors from each set, then $\mathbf{g}_{i,1}$ and these $k-1$ other vectors are linearly independent. This is possible since $\binom{n}{k-1}q^{k-1}<q^k$.
Write $\mathbf{s}_{i,j} = \sum_{l=1}^{j} \mathbf{g}_{i,l}$ for $j=1,\dots,r$.

Suppose we have $j$ vectors $\mathbf{g}_{i,1},\dots,\mathbf{g}_{i,j}$ such that when taken at most $k$ vectors from the sets $S_1, S_2, \dots, S_{i-1}$ or $\{\mathbf{g}_{i,1},\dots,\mathbf{g}_{i,j},\mathbf{s}_{i,j}\}$, with at most $r$ vectors from each set $S_1, S_2, \dots, S_{i-1}$ and at most $j$ vectors from the set $\{\mathbf{g}_{i,1},\dots,\mathbf{g}_{i,j},\mathbf{s}_{i,j}\}$, then these vectors are linearly independent.

Choose $\mathbf{g}_{i,j+1}$ to be any vector with the following two properties: When taken at most $k-1$ vectors from the sets $S_1, S_2, \dots, S_{i-1}$ or $\{\mathbf{g}_{i,1},\dots,\mathbf{g}_{i,j},\mathbf{s}_{i,j}\}$, with at most $r$ vectors from each set $S_1, S_2, \dots, S_{i-1}$ and at most $j$ vectors from the set $\{\mathbf{g}_{i,1},\dots,\mathbf{g}_{i,j},\mathbf{s}_{i,j}\}$, then $\mathbf{g}_{i,j+1}$ and these $k-1$ other vectors are linearly independent. Require also the following property: when taken at most $k-1$ vectors from the sets $S_1, S_2, \dots, S_{i-1}$ or $\{\mathbf{g}_{i,1},\dots,\mathbf{g}_{i,j},\mathbf{s}_{i,j}\}$, with at most $r$ vectors from each set $S_1, S_2, \dots, S_{i-1}$ and at most $j$ vectors from the set $\{\mathbf{g}_{i,1},\dots,\mathbf{g}_{i,j},\mathbf{s}_{i,j}\}$, then $\mathbf{s}_{i,j+1}$ and these $k-1$ other vectors are linearly independent. This is possible because there are at most $\binom{n}{k-1}$ different possibilities to choose, each of the options span a subspace with $q^{k-1}$ vectors, and since $q$ is large we have $2\binom{n}{k-1}q^{k-1}<q^k$. Notice that $\mathbf{s}_{i,j+1} \in V$ (where $V$ is some subspace) if and only if $\mathbf{g}_{i,j+1} \in \mathbf{-s}_{i,j}+V$.

To prove the induction step we have to prove the following thing: when taken at most $k-1$ vectors from sets $S_1, S_2, \dots, S_{i-1}$ or $\{\mathbf{g}_{i,1},\dots,\mathbf{g}_{i,j+1}\}$, with at most $r$ vectors from each set $S_1, S_2, \dots, S_{i-1}$ and at most $j$ vectors from the set $\{\mathbf{g}_{i,1},\dots,\mathbf{g}_{i,j+1}\}$, then $\mathbf{s}_{i,j+1}$ and these $k-1$ other vectors are linearly independent. Let $h \leq j$, $\mathbf{v}$ be a sum of at most $k-1-h$ vectors from the sets $S_1, S_2, \dots, S_{i-1}$ with at most $r$ vectors from each set, and let $m_1<\dots<m_h$ be indices in ascending order. We will assume a contrary: We have coefficients $c_{m_1},\dots,c_{m_h} \in \mathbf{F}_q$ such that
\begin{equation}
\mathbf{s}_{i,j+1} = \mathbf{v} + \sum_{l=1}^{h}c_{m_l}\mathbf{g}_{i,m_l}.
\end{equation}
If $m_h \neq j+1$ then our assumption is false by the definition so assume that $m_h = j+1$.
If $c_{j+1} \neq 1$ then
\begin{equation}
(1-c_{j+1})\mathbf{g}_{i,j+1} = \mathbf{v} + \sum_{l=1}^{h-1}c_{m_l}\mathbf{g}_{i,m_l} - \mathbf{s}_{i,j}.
\end{equation}
and again our assumption is false by the definition.
So assume that $c_{j+1} = 1$. Then we get
\begin{equation}
\mathbf{s}_{i,j} = \mathbf{v} + \sum_{l=1}^{h-1}c_{m_l}\mathbf{g}_{i,m_l}.
\end{equation}
and since $h-1 \leq j-1$ the assumption is false by the induction step.

Now, the sets $S_i$ consist of vectors $\{\mathbf{g}_{i,1},\dots,\mathbf{g}_{i,r},\mathbf{s}_{i,r}\}$ for $i=1,\dots,a$. If $b \neq 0$ the set $S_A$ consists of vectors $\{\mathbf{g}_{A,1},\dots,\mathbf{g}_{A,b-1},\mathbf{s}_{A,b-1}\}$. The matrix $\mathbf{G}$ is a matrix with vectors from the sets $S_1,S_2,\dots,S_A$ as its column vectors, \emph{i.e.},
\[
\mathbf{G}=\left(\mathbf{G}_1|\mathbf{G}_2|\dots|\mathbf{G}_A\right)
\]
where
\[
\mathbf{G_j}=\left(\mathbf{g}_{j,1}|\dots|\mathbf{g}_{j,r}|\mathbf{s}_{j,r}\right)
\]
for $i=1,\dots,a$, and
\[
\mathbf{G_A}=\left(\mathbf{g}_{A,1}|\dots|\mathbf{g}_{A,b-1}|\mathbf{s}_{A,b-1}\right)
\]
if $b\neq 0$.

To be a generator matrix for a code of dimension $k$ the rank of $\mathbf{G}$ has to be $k$. By the construction the rank is $k$ if and only if
$n-A \geq k$, and this is what we assumed.

\subsection{Lower Bound for the Largest Achievable Minimum Distance}\label{Subsec:Analysis}
In this subsection we will derive a lower bound for the largest achievable minimum distance of the linear codes with all-symbol locality. We will do this by analyzing the construction of Subsection \ref{Subsec:Construction}.

 Let $C$ be a linear code with a generator matrix $G$. A subset $A$ of the columns of $G$ is called a \emph{circuit} if $A$ is linearly dependent and all proper subsets of $A$ are linearly independent. A collection of circuits $C_1, \ldots, C_l$ of $C$ is called a \emph{nontrivial union} if
$$
C_i \nsubseteq \bigcup_{j \neq i} C_j, \hbox{ for  } 1 \leq i \leq l.
$$
To analyze our code construction we will use the following result that was proved by Tamo \emph{et al.} in \cite{LRCmatroid}.
\begin{thm}\label{Thm:mindistanceTamo}
The minimum distance of the linear locally repairable code is equal to
\[
d=n-k-\mu+2
\]
where $\mu$ is the minimum positive integer such that the size of every nontrivial union of $\mu$ circuits is at least $\mu+k$.
\end{thm}

To make the notations clearer we define $D_q(n,k,r)$ to be the minimum distance of our code construction for given parameters. To be exact, we have the following definition.
\begin{definition}
If our construction covers parameters $(n,k,r)$ over $\mathbb{F}_q$, then define $D_q(n,k,r)$ to be the minimum distance of such a code. If our construction does not cover parameters $(n,k,r)$ over $\mathbb{F}_q$, then define $D_q(n,k,r)$ to be zero.

\end{definition}

For the largest achievable minimum distance under the assumption of information symbol locality, we mark to be
\[
d_{\text{opt}}(n,k,r) := \max\left\{n-k-\left\lceil\frac{k}{r}\right\rceil+2 , 0 \right\}.
\]
The reason for this kind of definition is that if $n-k-\left\lceil\frac{k}{r}\right\rceil+2 \leq 0$ then it is impossible to have a code for parameters $(n,k,r)$.

Since the assumption of all-symbol locality is stronger than the assumption of information symbol locality, we know that
\begin{equation}\label{upperboundWithq}
D_q(n,k,r) \leq d_{\text{opt}}(n,k,r).
\end{equation}

In \cite{Gopalan} it was proved that there exists triplets $(n,k,r)$ such that the inequality \ref{upperboundWithq} is strict. So the natural question arises:
What is the relationship between $d_{\text{opt}}(n,k,r)$ and $D_q(n,k,r)$? Next we will study this question.

First we need a small straightforward lemma.
\begin{lemma}\label{lemma:kr}
Suppose $n-\left\lceil\frac{n}{r+1}\right\rceil \geq k$. Then $\frac{k}{r} \leq \frac{n}{r+1}$.
\end{lemma}

\begin{proposition}\label{Thm:construction}
Suppose $q>2\binom{n}{k-1}$, $k<n$, $n-\left\lceil\frac{n}{r+1}\right\rceil \geq k$, and $n \not\equiv 1 \mod r+1$. Then
\[
D_q(n,k,r)=d_{\text{opt}}(n,k,r)
\]
if $r+1$ divides $n$, and
\[
D_q(n,k,r) \geq n-k-\left\lfloor\frac{k}{r}-\frac{n}{r+1}\right\rfloor-\left\lfloor\frac{n}{r+1}\right\rfloor
\]
otherwise.
\end{proposition}
\begin{proof}
The construction of Subsection \ref{Subsec:Construction} gives a generating matrix $\mathbf{G}$ for a linear code. It is clear that the code it generates has the all-symbol repair locality $r$.

By Theorem \ref{Thm:mindistanceTamo} its minimum distance is $n-k-\mu+2$ where $\mu$ is a minimum positive integer $m$ with the following property: the size of every nontrivial union of $m$ circuits is at least $k+m$.

We remark that there are circuits of at most two types: possibly of size $k+1$ and those corresponding the sets $S_j$. Suppose we have a nontrivial union of $m$ circuits containing a circuit of size $k+1$. Then the size of this union is at least $(k+1)+(m-1)=k+m$.

Consider now only circuits corresponding the sets $S_j$. We have $A=\left\lceil\frac{n}{r+1}\right\rceil$ such circuits. It is easy to see that every union of such circuits is nontrivial. Write as before $n=a(r+1)+b$ with $0 \leq b < r+1$.

Suppose first that $b=0$. Then $|S_j|=r+1$ for all $j$. Each union of $m$ circuits has the same size
\[
|\cup_{j=1}^{m} S_{i_j}|=m(r+1)
\]
and $m(r+1) \geq m+k$ if and only if $m \geq \frac{k}{r}$, and hence $\mu=\min \left\{\left\lceil\frac{k}{r}\right\rceil ,A+1 \right\}= \left\lceil\frac{k}{r}\right\rceil$ by lemma \ref{lemma:kr}.

This gives that $D_q(n,k,r) =n-k-\left\lceil\frac{k}{r}\right\rceil+2$ when $r+1$ divides $n$.

Suppose now that $b\neq0$. Then $|S_j|=r+1$ for all $j$ except that $|S_A|=b$. Each minimal union of $m$ circuits contains the circuit corresponding the set $S_A$ and hence has the size
\[
|\cup_{j=1}^{m-1} S_{i_j} \cup S_A|=(m-1)(r+1)+b=mr-r+m-1+b.
\]
We have $mr-r+m-1+b \geq m+k$ if and only if
\[
m \geq \frac{k+1+r-b}{r} =1+\frac{k+1-n+\left\lfloor\frac{n}{r+1}\right\rfloor}{r}+\left\lfloor\frac{n}{r+1}\right\rfloor.
\]
Notice also that
\begin{equation}
\begin{split}
& \left\lceil 1+\frac{k+1-n+\left\lfloor\frac{n}{r+1}\right\rfloor}{r}+\left\lfloor\frac{n}{r+1}\right\rfloor \right\rceil \\

= & \left\lfloor\frac{k}{r}-\frac{n}{r+1}\right\rfloor+\left\lfloor\frac{n}{r+1}\right\rfloor+2 \\
\end{split}
\end{equation}
and hence
\begin{equation}
\begin{split}
\mu & =\min\left\{ \left\lfloor\frac{k}{r}-\frac{n}{r+1}\right\rfloor+\left\lfloor\frac{n}{r+1}\right\rfloor+2 , A+1 \right\} \\

& =\left\lfloor\frac{n}{r+1}\right\rfloor+2+\left\lfloor\frac{k}{r}-\frac{n}{r+1}\right\rfloor \\
\end{split}
\end{equation}
by lemma \ref{lemma:kr}.

This gives that
\[
D_q(n,k,r) \geq n-k-\mu+2 = n-k-\left\lfloor\frac{k}{r}-\frac{n}{r+1}\right\rfloor-\left\lfloor\frac{n}{r+1}\right\rfloor
\]
when $n \not\equiv 0,1 \mod r+1$.

\end{proof}

As a consequence the above analysis of the construction we have the following theorem.
\begin{thm}\label{Thm:MinDistanceSet}
Suppose $q>2\binom{n}{k-1}$ and $k<n$. Then $D_q(n,k,r) \in \{d_{\text{opt}}(n,k,r)-1,d_{\text{opt}}(n,k,r)\}$.
\end{thm}
\begin{proof}
Write $n=a(r+1)+b$ with $0 \leq b < r+1$.

Suppose first that
\begin{equation}\label{Eq:NotConstruction}
n-\left\lceil\frac{n}{r+1}\right\rceil +1 \leq k.
\end{equation}

If $r+1$ divides $n$ then the Equation \ref{Eq:NotConstruction} has the form
$
ar +1 \leq k
$
and hence
\[
n-k-\left\lceil \frac{k}{r}\right\rceil+2 \leq a(r+1)-(ar+1)-(a+1)+2=0.
\]
So it is impossible to have a code for parameters $(n,k,r)$ and hence $D_q(n,k,r)=d_{\text{opt}}(n,k,r)=0$.

If $r+1$ does not divide $n$ then the Equation \ref{Eq:NotConstruction} has the form
$
ar+b \leq k
$
and hence
\[
n-k-\left\lceil \frac{k}{r}\right\rceil+2 \leq a(r+1)+b-(ar+b)-(a+1)+2  \leq 1.
\]
Hence $D_q(n,k,r) \geq 0 \geq d_{\text{opt}}(n,k,r)-1$.

Suppose then that $n-\left\lceil\frac{n}{r+1}\right\rceil \geq k$. Now we can use Theorem \ref{Thm:construction}.

If $b=0$ then the claim is true by the Proposition \ref{Thm:construction}.

Assume $b=1$ and $\mathbf{G}$ is a generating matrix of a linear locally repairable code for parameters $(n-1,k,r)$ and minimum distance $d_{\text{opt}}(n-1,k,r)$. Replicate any column in $\mathbf{G}$ and get a generating matrix for a linear locally repairable code for parameters $(n,k,r)$ and minimum distance $d_{\text{opt}}(n,k,r)-1$.

Assume $b>1$. Then
\[
D_q(n,k,r) \geq n-k-\left\lfloor\frac{k}{r}-\frac{n}{r+1}\right\rfloor-\left\lfloor\frac{n}{r+1}\right\rfloor
\]
and hence
\begin{equation}\label{Eq:optimijaalaraja}
\begin{split}
& d_{\text{opt}}(n,k,r) - D_q(n,k,r) \\
\leq & \left\lfloor\frac{k}{r}-\frac{n}{r+1}\right\rfloor -\left\lceil\frac{k}{r}\right\rceil + \left\lfloor\frac{n}{r+1}\right\rfloor +2 \\

\leq & \left\lfloor-\frac{n}{r+1}\right\rfloor + \left\lfloor\frac{n}{r+1}\right\rfloor +2 =  -1 +2 =1.\\
\end{split}
\end{equation}

\end{proof}

So we know that if $d_{\text{opt}}(n,k,r) \geq 2$ then we have a linear locally repairable code for parameters $(n,k,r)$. If $d_{\text{opt}}(n,k,r) = 0$ then it is impossible to have a linear locally repairable code for parameters $(n,k,r)$. However, if $d_{\text{opt}}(n,k,r) = 1$ then we do not know whether there exists a linear locally repairable code for parameters $(n,k,r)$.

Theorem \ref{Thm:MinDistanceSet} gives a lower bound for the minimum distance. In fact we can say little more in a certain case.

Below, the \emph{fractional part} of of $x$ is denoted by $\{x\}$, \emph{i.e.},  $\{x\} = x - \lfloor x\rfloor$.
\begin{thm}\label{Thm:MinDistanceWithFracPart}
Suppose $q>2\binom{n}{k-1}$, $k<n$, $\left\{\frac{k}{r}\right\} < \left\{\frac{n}{r+1}\right\}$, and $r$ does not divide $k$. Then
\[
D_q(n,k,r)=d_{\text{opt}}(n,k,r).
\]
\end{thm}
\begin{proof}
Write $n=a(r+1)+b$ with $0 \leq b < r+1$. If $b= 0 \text{ or } 1$ then $\left\{\frac{k}{r}\right\} \geq \left\{\frac{n}{r+1}\right\}$ so we may assume that this is not the case.

Suppose first that $n-\left\lceil\frac{n}{r+1}\right\rceil \geq k$. By studying the Equation \ref{Eq:optimijaalaraja} again we notice that
\begin{equation}\label{Eq:optimijaalarajaTarkemmin}
\begin{split}
& d_{\text{opt}}(n,k,r) - D_q(n,k,r) \\
\leq & \left\lfloor\frac{k}{r}-\frac{n}{r+1}\right\rfloor -\left\lceil\frac{k}{r}\right\rceil + \left\lfloor\frac{n}{r+1}\right\rfloor +2 \\

= & \left\lfloor\left\{\frac{k}{r}\right\}-\left\{\frac{n}{r+1}\right\}\right\rfloor +1 =0 \\
\end{split}
\end{equation}

Suppose then that $n-\left\lceil\frac{n}{r+1}\right\rceil +1 \leq k$. Since $\left\{\frac{k}{r}\right\} < \left\{\frac{n}{r+1}\right\}$ we know that $r+1$ cannot divide $n$ and hence we have $ar+b \leq k$. It is impossible that $ar+b = k$. Indeed, then we would have $\left\{\frac{k}{r}\right\}=\left\{\frac{b}{r}\right\} > \left\{\frac{b}{r+1}\right\}=\left\{\frac{n}{r+1}\right\}$. Hence $ar+b < k$ and
\begin{equation}
\begin{split}
d_{\text{opt}}(n,k,r) & = \max\left\{n-k-\left\lceil\frac{k}{r}\right\rceil+2,0\right\} \\
& \leq \max\left\{1-\left\lceil\frac{b+1}{r}\right\rceil,0\right\}=0
\end{split}
\end{equation}
and hence $D_q(n,k,r)=d_{\text{opt}}(n,k,r)$.
\end{proof}

\section{Constructing Optimal LRCs over $\mathbb{F}_4$} \label{Subsec:Construction-Optimal}

In this section we give some constructions of optimal LRCs over the field of four elements $\mathbb{F}_4$ for certain values of $(r,d)$. Our LRCs will be described in the setting of matrices with different operators as entries.

\subsection{Matrix Representation}

We represent the elements of $\mathbb{F}_4$ as $\{00,01,10,11\}$, such that the addition of elements in $\mathbb{F}_4$ can be considered as bitwise addition without carry (e.g. $01 + 11 = 10$). In our construction of optimal LRCs over $\mathbb{F}_4$ we will use the operators $\alpha$, $\alpha^2$, $\beta$, $\beta^2$, $\beta^3$, $1 \hbox{ and } 0$ on $\mathbb{F}_4$ to $\mathbb{F}_4$ defined as
$$
\begin{array}{rrrr}
\alpha(00) = 00, & \alpha(01) = 10, & \alpha(10) = 11 & \alpha(11) = 01,\\
\alpha^2(00) = 00, & \alpha^2(01) = 11, & \alpha^2(10) = 01 & \alpha^2(11) = 10,\\
\beta(00) = 01, & \beta(01) = 10, & \beta(10) = 11 & \beta(11) = 00,\\
\beta^2(00) = 10, & \beta^2(01) = 11, & \beta^2(10) = 00 & \beta^2(11) = 01,\\
\beta^3(00) = 11, & \beta^3(01) = 00, & \beta^3(10) = 01 & \beta^3(11) = 10,\\
1(00) = 00, & 1(01) = 01, & 1(10) = 10 & 1(11) = 11,\\
0(00) = 00, & 0(01) = 00, & 0(10) = 00 & 0(11) = 00.
\end{array}
$$

A code $C$ is represented by a $k \times n$ matrix $F$. The entries of the matrix are the operators $\alpha, \alpha^2, \beta, \beta^2, \beta^3, 1 \hbox{ and } 0$. The code $C$ consists of the following codewords
$$
C = \{\boldsymbol{y} \in \mathbb{F}_4^n : y_j = F_{1,j}(x_1) + \ldots + F_{k,j}(x_k) \hbox{ for } \boldsymbol{x} \in \mathbb{F}_4^k\}.
$$

\subsection{Optimal LRCs over $\mathbb{F}_4$}

Let $A$ and $B$ be the following matrices
$$
A = \left (
\begin{array}{cccc}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1
\end{array}
\right )
\quad \hbox{and} \quad
B = \left (
\begin{array}{ccc}
0 & 1 & 1 \\
0 & \alpha & \alpha \\
0 & \alpha^2 & \alpha^2
\end{array}
\right ).
$$
For $i \geq 1$, let $F_i^1(3,3)$ be the $(i+1) \times (i+1)$-block matrix
$$
\begin{small}
F_i^1(3,3) =
\left (
\begin{array}{c | c | c | ccc}
 A & & & & B   & \\
\hline
& \ddots & & & \vdots & \\
\hline
& & A  &  & B & \\
\hline
&&& 1 & \alpha & \alpha^2
\end{array}
\right ),
\end{small}
$$
where the entires of the empty blocks are 0-operators and the first $i$ diagonal blocks are $A$-blocks.

\begin{thm} \label{th:linear-optimal-(3,3)}
The matrix $F_i^1(3,3)$ defines a locally repairable code $C$ over $\mathbb{F}_4$ with parameters $(n,k,d,r) = (4i+3, 3i+1,3,3)$ for $i \geq 1$.
\end{thm}
\begin{proof}
Let $f:\mathbb{F}_4^{3i+1} \rightarrow \mathbb{F}_4^{4i+3}$ denote the mapping given by the matrix $F_i^1(3,3)$. Now, $f$ is injective, because
\begin{equation} \label{eq:identity_map_in_f}
\begin{array}{l}
y_{4j-3} = x_{3j-2}, y_{4j-2} = x_{3j-1}, y_{4j-1} = x_{3j} \hbox{ and}\\
y_{4i+1} = x_{3i+1},
\end{array}
\end{equation}
for $1 \leq j \leq i$ and $f(\boldsymbol{x}) = \boldsymbol{y}$. Since $f$ is injective and $F_i^1(3,3)$ is a $(3i+1) \times (4i+3)$-matrix it follows that $(n,k) = (4i+3,3i+1)$.

The code $C$ has repair locality $r=3$, since from the fact that $1(x) + \alpha(x) + \alpha^2(x) = 00$ for $x \in \mathbb{F}_4$ we may deduce that
$$
y_{4i+1} + y_{4i+2} + y_{4i+3} = 00,
$$
for $f(\boldsymbol{x}) = \boldsymbol{y}$. Moreover, we have that
$$
y_{4j-3} + y_{4j-2} + y_{4j-1} + y_{4j} = 00,
$$
for $f(\boldsymbol{x}) = \boldsymbol{y}$ and $1 \leq j \leq i$.

Let $d(\boldsymbol{u}, \boldsymbol{v})$ denote the distance between $\boldsymbol{u}, \boldsymbol{v} \in \mathbb{F}_4^m$.
Suppose $\boldsymbol{w}$ and $\boldsymbol{x}$ are two elements of $\mathbb{F}_4^{3i+1}$ such that $d(\boldsymbol{w}, \boldsymbol{x}) \geq 3$. Then, by (\ref{eq:identity_map_in_f}), we deduce that $d(f(\boldsymbol{w}), f(\boldsymbol{x})) \geq 3$.

Suppose $\boldsymbol{w}$ and $\boldsymbol{x}$ are two elements of $\mathbb{F}_4^{3i+1}$ such that $d(\boldsymbol{w}, \boldsymbol{x}) = 1$. We note that every row of $F_i^1(3,3)$ has at least three entries $a$, $b$ and $c$ with operators  1, $\alpha$ or $\alpha^2$. In particular, this yields that the coefficients $a$, $b$ and $c$ of $f(\boldsymbol{w})$ differ from these coefficients of $f(\boldsymbol{x})$, and hence $d(f(\boldsymbol{w}), f(\boldsymbol{x})) \geq 3$.

Suppose $\boldsymbol{w}$ and $\boldsymbol{x}$ are two elements of $\mathbb{F}_4^{3i+1}$ such that $d(\boldsymbol{w}, \boldsymbol{x}) = 2$. Let $a$ and $b$ be the index of the two coefficients in which $\boldsymbol{w}$ and $\boldsymbol{x}$ differ. Assume that row $a$ and row $b$ of $F_i^1(3,3)$ are in different horizontal blocks. Then there are at least three columns $e$, $g$ and $h$ of $F_i^1(3,3)$ such that one of the entries $(a,e)$ and $(b,e)$ is the 0-operator and the other one is the 1-operator, this property also holds for the entries $(a,g)$, $(b,g)$ and $(a,h)$, $(b,h)$. Consequently, $d(f(\boldsymbol{w}), f(\boldsymbol{x})) \geq 3$ when the rows $a$ and $b$ are in different horizontal blocks.

Now, suppose that row $a$ and row $b$ are in the same horizontal block of $F_i^1(3,3)$, i.e. row $a$ and $b$ are rows in a submatrix of the following form
$$
\left (
\begin{array}{c|c|c|c}
 \mathbf{0}& A & \mathbf{0} & B
 \end{array}
 \right).
$$
\begin{comment}
$$
\left (
\begin{array}{ccc|cccc|ccc|ccc}
0&\ldots&0&1&0&0&1&0&\ldots&0&0&1&1\\
0&\ldots&0&0&1&0&1&0&\ldots&0&0&\alpha&\alpha\\
0&\ldots&0&0&0&1&1&0&\ldots&0&0&\alpha^2&\alpha^2
\end{array}
\right ).
$$
\end{comment}
It is easy to check by hand that if
$$
y \neq y' \hbox{, } z \neq z' \hbox{ and } 1(y) + 1(z) = 1(y') + 1(z'),
$$
then
$$
\begin{array}{l}
1(y) + \alpha(z) \neq 1(y') + \alpha(z'),\\
1(y) + \alpha^2(z) \neq 1(y') + \alpha^2(z'),\\
\alpha(y) + \alpha^2(z) \neq \alpha(y') + \alpha^2(z').
\end{array}
$$
for $y,y',z,z' \in \mathbb{F}_4$. As a consequence of this fact and since there are two pair of entries $\{(a,g),(b,g)\}$ and $\{(a,h),(b,h)\}$ in $F_i^1(3,3)$ such that one of the entries in each pair is the 0-operator and the other entry is the 1-operator, we deduce that $d(f(\boldsymbol{w}), f(\boldsymbol{x})) \geq 3$. Hence $d \geq 3$ for the code.

Moreover, since
$$
4i + 3 - (3i + 1) - \left \lceil \frac{3i+1}{3} \right \rceil + 2 = 3
$$
we obtain that $C$ is an optimal $(4i+3,3i+1,3,3)$-LRC.
\end{proof}

Let $D$ be the following matrix
$$
D = \left (
\begin{array}{cccc}
0 & 0 & 1 & 1 \\
0 & 0 & \alpha & \alpha \\
0 & 0 & \alpha^2 & \alpha^2
\end{array}
\right ).
$$
For $i \geq 1$, let $F_i^2(3,3)$ be the $(i+1) \times (i+1)$-block matrix
$$
F_i^2(3,3) =
\left (
\begin{array}{c | c | c |cccc}
 A & & & & D   & &\\
\hline
& \ddots & & & \vdots & & \\
\hline
& & A  &  & D & & \\
\hline
&&& 1 & 0 & \beta & \beta^2\\
&&& 0 & 1 & \beta^2 & \beta
\end{array}
\right ),
$$
and let $F_i^1(3,4)$ be the $(i+1) \times (i+1)$-block matrix
$$
F_i^1(3,4) =
\left (
\begin{array}{c | c | c |cccc}
 A & & & & A   & &\\
\hline
& \ddots & & & \vdots & & \\
\hline
& & A  &  & A & & \\
\hline
&&& 1 & \beta & \beta^2 & \beta^3
\end{array}
\right ).
$$

With similar proof techniques as in Theorem \ref{th:linear-optimal-(3,3)} we can prove the following two theorems.

\begin{thm} \label{th:nonlinear-optimal-(3,3)}
The matrix $F_i^2(3,3)$ defines a locally repairable code $C$ over $\mathbb{F}_4$ with parameters $(n,k,d,r) = (4i+4, 3i+2,3,3)$ for $i \geq 1$.
\end{thm}

\begin{thm} \label{th:nonlinear-optimal-(3,4)}
The matrix $F_i^1(3,4)$ defines a locally repairable code $C$ over $\mathbb{F}_4$ with parameters $(n,k,d,r) = (4i+4, 3i+1,3,4)$ for $i \geq 1$.
\end{thm}

Note that the codes we construct in Theorem \ref{th:nonlinear-optimal-(3,3)} and Theorem \ref{th:nonlinear-optimal-(3,4)} are nonlinear since $\beta$ is a nonlinear operator over $\mathbb{F}_4$.

\section{Future Work}

As future work it is still left to find the exact expression of the largest achievable minimum distance of a linear locally repairable code with all-symbol locality when given the length $n$, dimension $k$, and locality $r$ of the code. Our goal is to also generalize the constructions given in Section \ref{Subsec:Construction-Optimal} to other parameters $r$ and $d$ over small fields.

config: arxiv


TITLE: calculate structure automorphism group in MAGMA
QUESTION [2 upvotes]: Please hint me! Is there any instruction in MAGMA to calculate structure automorphism group by using generators?
for example:
> G:=Graph<16| { 1, 2 }, { 1, 3 }, { 1, 14 }, { 2, 5 }, { 2, 7 }, { 3, 4 }, { 3, 14 },
> { 4, 6 }, { 4, 12 }, { 5, 6 }, { 5, 10 }, { 6, 12 }, { 7, 8 }, { 7, 15 },
> { 8, 9 }, { 8, 15 }, { 9, 10 }, { 9, 11 }, { 10, 11 }, { 11, 13 },
> { 12, 13 }, { 13, 16 }, { 14, 16 }, { 15, 16 }    >;

>  A:=AutomorphismGroup(G);

> Generators(A);

{
    (1, 7)(3, 8)(4, 9)(6, 10)(11, 12)(14, 15),
    (1, 6)(2, 5)(3, 4)(7, 10)(8, 9)(11, 15)(12, 14)(13, 16),
    (1, 11)(2, 13)(3, 9)(4, 8)(5, 16)(6, 15)(7, 12)(10, 14)
}

I want to calculate structure of A?
in GAP we have "StructureDescription" to calculate structure of  automorphism group:
gap> A:=Group([(1,7)(3,8)(4,9)(6,10)(11,12)(14,15),
> (1,6)(2,5)(3,4)(7,10)(8,9)(11,15)(12,14)(13,16),
> (1,11)(2,13)(3,9)(4,8)(5,16)(6,15)(7,12)(10,14)]);

Group([ (1,7)(3,8)(4,9)(6,10)(11,12)(14,15),
(1,6)(2,5)(3,4)(7,10)(8,9)(11,15)(12,14)(13,16),
(1,11)(2,13)(3,9)(4,8)(5,16)(6,15)(7,12)(10,14) ])

gap> StructureDescription(A);

"C2 x C2 x C2"

REPLY [1 votes]: If you have both Magma and GAP at your disposal and if the group in question is not of too large an order, there is a neat little trick I've used a couple of times.
Given a group $G$ in Magma, type the following.
IdentifyGroup(G);

This will give you two numbers $a,b$ which identifies the group in the Small Groups Library.
Then you can start GAP and type this:
G:=SmallGroup(a,b);
StructureDescription(G);


text:
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TITLE: How to prove that $e^x=x$ has no real solution?
QUESTION [2 upvotes]: I am really stumped by this equation.
$$e^x=x$$
I need to prove that this equation has no real roots. But I have no idea how to start.
If you look at the graphs of $y=e^x$ and $y=x$, you can see that those graphs do not meet anywhere. But I am trying to find an algebraic and rigorous proof. Any help is appreciated. ( $e^\infty=\infty$ looks like a solution but is it?)

REPLY [1 votes]: We have $e^x=x$ if and only if $x=\ln x$, which means, for one thing that there can be no solution with $x\le0$.  For $x\gt0$, we have
$$\ln x=\int_1^x{dt\over t}\le\int_1^xdt=x-1\lt x$$
so there is no solution for $x\gt0$ either.
Remark:  The key inequality between the two integerals may look puzzling for $x\lt1$, since ${1\over t}\gt1$ in that case; but it's still correct because the integrals are both negative.


text:
\subsection{Single Patch Integration}

The following test investigates the impact of the reduced number of geometry evaluations to the calculation time with respect to the subparametric formulation. Such operations are needed for the calculation of tangents and outward normals to the NURBS patch. The test setting is illustrated in \figref{fig:integration_test_setting}. A linear, flat NURBS patch $\be$ is placed in the $xy$--plane. Three different collocation points are chosen, so that the boundary integral gets singular, nearly singular and regular and thus subject to different integration strategies as described in \secref{sec:integration}. The error of integration is chosen $\err_Q=10^{-9}$. The integration is performed for the discrete single layer $\mat{V}_\be$ and double layer potential $\mat{K}_\be$. For accurate integration, a significant amount of Gauss points are needed. Especially for the singular integration, the amount increases rapidly. This is demonstrated in \tabref{tab:singlePatchIntegration_superparametrix} for this particular case.

It is placed in the xy--plane and three collocation points are defined.
The first is a corner node of the square which causes singular integration.
The second is very close to the square which yields to nearly singular integration and for the third is far enough away so that a regular integration can be performed.
The test setting is illustrated in \figref{fig:integration_test_setting}.

\tikzfig{tikz/integration_test_setting}{1.5}
{
	Position of the unit square and the collocation points, marked by black dots, for the regular, nearly singular and singular cases.
}{integration_test_setting}

For each collocation point the single and double layer operator are applied to the NURBS patch.
The numerical integration is performed by the adaptive scheme described in \secref{sec:integration}, with an maximum allowed error of $\err_Q=10^{-9}$.
The resulting number of Gauss point for the initial integrand ($p=1$) are listed in  \tabref{tab:singlePatchIntegration_superparametrix}.

\begin{mytable}
{The number of Gauss points used for the numerical integration of the initial and superparametric patch.}
{singlePatchIntegration_superparametrix}
{lcccc}
\mytableheader{
	Integration			& Kernel U 			& Kernel T}
	regular  	& 64					& 64		\\
	nearly singular  	& 375		     	& 415	\\
	singular  	& 1447 			         & 1015 	\\
\end{mytable}

The initial representation is refined by order elevation up to $p=6$.
Thereby, three different patch types are analyzed.
The superparametric case, where only the basis of the geometry is order elevated, the subparametric case, where only the basis of unknowns is order elevated and the isoparametric case, where the basis of both is order elevated.
Note that the integrand change if the basis of unknowns is refined. However, the norm of the resulting vector is only influence by the distance of the collocation point.

For each patch type and order, the numerical integration is performed 100 times and the total integration time is measured.
In \figref{fig:time_super}, the integration time in the superparametric case $t_{super}$ is related to the integration time of the initial unrefined patch $t_{init}$.
The ratio of the integration time on the conventional isoparametric $t_{iso}$ and the proposed subparametric $t_{sub}$ patch is shown in \figref{fig:time_sub}.\mycomment[BM]{Schrift skaliert mit Bildgröße - groupplot Problem? }

\tikzfig{plots/GP-plane_planestrain_KERNEL_U_and_T_GEOMETRY}{1.0}
{
	The integration time of the superparametric patch ($t_{super}$) related to the initial patch ($t_{init}$) for the regular, nearly singular and singular integral with the U kernel (left) and T kernel (right), respectively.
	The order of the basis functions related to the geometry are denoted by $p$.

}{time_super}

\tikzfig{plots/GP-plane_planestrain_KERNEL_U_and_T_ALL}{1.0}
{
	The integration time of the isoparametric ($t_{iso}$) related to the subparametric ($t_{sub}$) patch for U kernel (left) and T kernel (right).
	The order $p$ corresponds to the refined basis functions.
}{time_sub}

Both time comparisons verify the increasing numerical effort due to the order elevation of the geometry.
It is worth noting that there is hardly any difference between the regular, nearly singular and singular integration, although the number of Gauss points rises rapidly if the collocation point comes closer to the patch.
This indicates that the governing factor of the relative integration time is the efficiency of the geometry evaluation.
The benefit of the subparametric patch is for the Kernel T slightly better because the evaluation of the Kernel includes the outward normal of the surface.
However, note that accurate integration requires an immense quantity of Gauss points, which leads to a high number of evaluations of geometry related informations.