text:
\section{Introduction}

Given a knot $K$ in $S^3$ and an $r\in \mathbf{Q}\cup \{\infty\},$ we denote the \textit{Dehn surgery} on $K$ with slope $r$ by $S^3_r(K).$  Surgeries on $K$ along distinct slopes $r$ and $r'$ are called \textit{cosmetic} if $S^3_r(K)$ and $S^3_{r'}(K)$ are homeomorphic manifolds.  Furthermore, a pair of such surgeries is said to be \textit{purely cosmetic} if $S^3_r(K)$ and $S^3_{r'}(K)$ are homeomorphic as \textit{oriented} manifolds.  We use the symbol $\cong$ to denote ``orientation-preserving homeomorphic." If, on the other hand, $S^3_r(K) \cong -S^3_{r'}(K),$ we say this pair of surgeries is \textit{chirally cosmetic}; here $-M$ denotes the manifold $M$ with reversed orientation.

No purely cosmetic surgeries are have been found on nontrivial knots in $S^3$; indeed, Problem 1.81(A) in \cite{KirbyList} conjectures that none exist.  On the other hand, there are examples of chirally cosmetic surgeries.  For instance, $S^3_r(K) \cong -S^3_{-r}(K)$ whenever $K$ is an amphicheiral knot.  Also, $(2,n)$-torus knots are known to admit chirally cosmetic surgeries; see \cite{IIS,Mat}.

In this work, we consider the family of \textit{alternating odd pretzel knots}.  These are pretzel knots of the form $P(2k_1+1,2k_2+1,\dots,2k_{2g+1}+1)$, where the $k_i$ are integers all with the same sign; hence, up to mirror image, we may assume all $k_i$ are nonnegative.  See Figure \ref{fig:Pknot} for the five-stranded case.  It is a fact that $g$ is the Seifert genus of the knot.  For convenience, we will often use the shorthand $K(k_1,k_2,\dots,k_{2g+1})$ to denote these knots.

\begin{figure}
\centering
\includegraphics[width=.75\textwidth]{pretzel}
\caption{The pretzel knot $P(2k_1+1,2k_2+1,2k_3+1,2k_4+1,2k_5+1).$  Here the boxes represent $k_i$ full right-handed twists of the strands passing through.}
\label{fig:Pknot}
\end{figure}

Note that these knots are already known not to admit any purely cosmetic surgeries either.  This can be seen, for instance, from the fact that of the signatures and $v_3$ of these knots are nonzero(see section \ref{sec:comp} below).  By \cite{Hans}, alternating knots that admit purely cosmetic surgeries must have zero signature.  Alternatively, by \cite{IW}, knots admitting purely cosmetic surgeries must have $v_3=0.$

Our first main result is that almost none of the alternating odd pretzel knots of genus 2 and 3 admit chirally cosmetic surgeries.

\begin{restatable}{thm}{main}\label{thm:main}
\begin{enumerate}[label=(\roman*)]
\item Let $K=K(k_1,k_2,k_3,k_4,k_5)$ with each $k_i \geq 0.$  If at least one $k_i>0,$ then $K$ does not admit any chirally cosmetic surgeries.
\item Let $K=K(k_1,k_2,k_3,k_4,k_5,k_6,k_7)$ with each $k_i \geq 0.$  If at least one $k_i>0,$ then $K$ does not admit any chirally cosmetic surgeries.
\end{enumerate}
\end{restatable}
\begin{proof}[Remark.]\let\qed\relax
$K(0,0,0,0,0)$ is the (-2,5)-torus knot, and $K(0,0,0,0,0,0,0)$ is the (-2,7)-torus knot. Both of these are already known to admit (infinitely many pairs of) chirally cosmetic surgeries; see \cite{Mat} and \cite{IIS}.
\end{proof}
Ichihara, Ito, and Saito have already shown the analogue of this for the genus 1 (three-stranded) case.  Indeed, they classified chirally cosmetic surgeries for all alternating genus 1 knots \cite{IIS}.

For the case of higher genus, we show more generally that, given a fixed genus, all but finitely many alternating odd pretzel knots admit no chirally cosmetic surgeries.  More precisely, we have the following:

\begin{restatable}{thm}{gen}\label{thm:gen}
Let $K=K(k_1,\dots,k_{2g+1})$.  If $k_1+\dots+k_{2g+1} \geq \alpha g$ then $K$ does not admit any chirally cosmetic surgeries.  Here $\alpha = \frac{9+\sqrt{237}}{12} \approx 2.0329$.
\end{restatable}
\begin{proof}[Note:]\let\qed\relax
The crossing number of $K(k_1,\dots,k_{2g+1})$ is $2(k_1+\dots+k_{2g+1})+2g+1$ and so one may equivalently rephrase the theorem as saying that such knots admit no chirally cosmetic surgeries whenever the crossing number is at least $2(\alpha+1)g+1$.
\end{proof}

\section{Finite Type Invariants}
Here we briefly recall some facts about finite type invariants (also called Vassiliev invariants) for knots.  Suppose a knot invariant $v$ can be extended to an invariant of \textit{singular knots} (i.e., knots with possibly finitely many points of self-intersection) in a way that satisfies the following:
\[
v(K_{\bullet}) = v(K_+)-v(K_-)
\]
whenever the (singular) knots $K_+$, $K_-$, and $K_{\bullet}$ differ locally near a crossing/self-intersection as in Figure \ref{fig:skein}.  Then $v$ is said to be a finite type invariant of order $n$ if moreover $v(K)=0$ whenever $K$ has at least $n+1$ self-intersection points.

\begin{figure}
\centering
\includegraphics[width=.75\textwidth]{crossing_figure4}
\caption{The diagrams of $K_+$, $K_-$, $K'\cup K''$, and $K_{\bullet}$ differ only locally near a crossing.}
\label{fig:skein}
\end{figure}

For example, recall the Conway polynomial of a knot, which is related to the Alexander polynomial in the following way:
\[
\nabla_K(z) = \Delta_K(t)|_{z=t^{1/2}-t^{-1/2}}
\]
For a knot $K$, the Conway polynomial will have the form:
\[
\nabla_K(z) = 1 + \sum_{j=1}^{n}a_{2j}(K)z^{2j}
\]
It is a fact that the coefficent $a_{2j}(K)$ is a finite type invariant of order $2j$ for each $j$.

The other finite type invariant that will be of interest to us is $v_3$.  This is a third-order invariant, which may be defined as:
\[
v_3(K) = -\frac{1}{36}V'''_{K}(1) -\frac{1}{12}V''_{K}(1)
\]
where $V_K(t)$ is the Jones polynomial of $K$.
\subsection{Computing the Invariants for Pretzel Knots}\label{sec:comp}
We consider the case $K=K(k_1,k_2,k_3,k_4,k_5)$ first.  Applying Seifert's algorithm to a diagram such as the one in Figure \ref{fig:Pknot}, one finds that the Seifert form for $K$ may be given by the following matrix:
\[
A=\left(\begin{matrix}
k_1+k_2+1 & k_2 & 0 & 0 \\
k_2+1 & k_2+k_3+1 & k_3 & 0 \\
0 & k_3+1 & k_3+k_4+1 & k_4 \\
0 & 0 & k_4+1 & k_4+k_5+1
\end{matrix}\right)
\]
From this, we may compute the Alexander and Conway polynomials.  In particular, $\Delta_K(t)=\det\left(t^{1/2}A-t^{-1/2}A^T\right)$ and $\nabla_K(z) = \Delta_K(t)|_{z=t^{1/2}-t^{-1/2}} = 1+ a_2(K)z^2+a_4(K)z^4$.  Explicit computation gives:
\begin{equation}\label{eq:a242}
\begin{cases}
a_2(K) = 3 + 2 s_{1,5} + s_{2,5}\\
a_4(K) = 1 + s_{1,5} + s_{2,5} + s_{3,5} + s_{4,5}
\end{cases}
\end{equation}
Here we use $s_{n,m}$ as a shorthand for $s_n(k_1,\dots,k_m),$ the $n$th elementary symmetric polynomial in $k_1,\dots,k_m$ given by:
\[
s_{n,m}=\sum_{\substack{P\subseteq\{1,\dots,m\} \\ |P|=n}} \prod_{j\in P} k_j
\]
Let us here record some properties of the elementary symmetric polynomials, which are straightforward consequences of their definition.
\begin{lem}\label{lem:sym}
Let $s_{n,m}$ denote the $n$th elementary symmetric polynomial in $k_1,\dots,k_m$.  Then, the following hold:
\begin{itemize}
\item If $k_j=0$ for all $j>N$, then whenever $n\geq N$, $s_{n,m} = s_{n,N}$
\item $s_{n,m+1} = s_{n,m} + k_{m+1}s_{n-1,m}$
\end{itemize}
\end{lem}
\begin{proof}[Remark.]
By convention, we take $s_{0,m}=1$, and if $n>m$, $s_{n.m}=0$.
\end{proof}

Now we turn to the case $K=K(k_1,k_2,k_3,k_4,k_5,k_6,k_7)$.  As before, we can represtent the Seifert form with the following matrix:
\[
A=\left(\begin{matrix}
k_1+k_2+1 & k_2 & 0 & 0 & 0 & 0 \\
k_2+1 & k_2+k_3+1 & k_3 & 0 & 0 & 0 \\
0 & k_3+1 & k_3+k_4+1 & k_4 & 0 & 0 \\
0 & 0 & k_4+1 & k_4+k_5+1 & k_5 & 0 \\
0 & 0 & 0 & k_5+1 & k_5+k_6+1 & k_6 \\
0 & 0 & 0 & 0 & k_6+1 & k_6+k_7+1
\end{matrix}\right)
\]
Once again, computation of the Conway polynomial gives:
\begin{equation}\label{eq:a243}
\begin{cases}
a_2(K) = 6 + 3 s_{1,7} + s_{2,7}\\
a_4(K) = 5 + 4s_{1,5} + 3s_{2,5} + 2s_{3,5} + s_{4,5}\\
a_6(K) = 1 + s_{1,7} + s_{2,7} + s_{3,7} + s_{4,7} + s_{5,7} + s_{6,7}
\end{cases}
\end{equation}
In the general case where $K=K(k_1,\dots,k_{2g+1})$, we will need the following formulae for $a_2(K)$ and $v_3(K)$.
\begin{lem}\label{lem:av}
Let $K=K(k_1,\dots,k_{2g+1})$.  Then
\begin{align*}
a_2(K)&=\frac{1}{2}g(g+1) + g s_{1,2g+1} + s_{2,2g+1}
\\
v_3(K)&=-\frac{1}{2}\left(\frac{g(g+1)(2g+1)}{3} + g(2g+1)s_{1,2g+1} + g s_{1,2g+1}^2 + 2g s_{2,2g+1} + s_{1,2g+1}s_{2,2g+1} + s_{3,2g+1} \right)
\end{align*}
\end{lem}
\begin{proof}
We employ the following skein relation for $a_2$ \cite{IW}:
\begin{equation}\label{eq:skein2}
a_2(K_+)-a_2(K_-)=\mathrm{lk}(K',K'')
\end{equation}
Here $K_+$ and $K_-$ differ at a single crossing, and $K'$ and $K''$ are the two components of the link that results from the oriented resolution at that crossing; see Figure \ref{fig:skein} for an illustration.

We proceed by induction on $g$.  Notice that $g=0$ corresponds to the unknot, which has $a_2=0$.
Now consider $g \geq 1$
Considering one of the crossings on the $2g+1$st strand, we have $K_{-}=K(k_1,\dots,k_{2g+1})$, $K_{+}=K(k_1,\dots,k_{2g+1}-1),$ and $K' \cup K''$ is a pair of unknots with linking number $-(k_1+\dots+k_{2g}+g) = -(s_{1,2g}+g)$.  Applying (\ref{eq:skein2}) gives:
\[
a_2(K(k_1,\dots,k_{2g+1}))-a_2(K(k_1,\dots,k_{2g+1}-1)) = s_{1,2g}+g
\]
Repeating this procedure $k_{2g+1}-1$  times gives:
\[
a_2(K(k_1,\dots,k_{2g+1}))-a_2(K(k_1,\dots,k_{2g},0)) = k_{2g+1}(s_{1,2g}+g)
\]
Now repeating this procedure for the $2g$th strand gives:
\begin{multline}\label{eq:ind2}
a_2(K(k_1,\dots,k_{2g+1}))-a_2(K(k_1,\dots,k_{2g-1},0,0)) \\= k_{2g}(s_{1,2g-1}+g) + k_{2g+1}(s_{1,2g}+g)
\end{multline}
Now, applying (\ref{eq:skein2}) to one of the two rightmost crossings of $K(k_1\dots,k_{2g-1},0,0)$, we see that:
\[
a_2(K(k_1,\dots,k_{2g-1},0,0)) - a_2(K(k_1,\dots,k_{2g-1})) = (s_{1,2g-1}+g)
\]
Combining this with (\ref{eq:ind2}) and applying the induction hypothesis, we find:
\begin{align*}
a_2(K&(k_1,\dots,k_{2g+1})) = a_2(K(k_1,\dots,k_{2g-1}))  + (k_{2g}+1)(s_{1,2g-1}+g) + k_{2g+1}(s_{1,2g}+g) \\
&= \frac{1}{2}g(g-1) + (g-1) s_{1,2g-1} + s_{2,2g-1} + (k_{2g}+1)(s_{1,2g-1}+g) + k_{2g+1}(s_{1,2g}+g)\\
&= \frac{1}{2}g(g-1) + g + (g-1) s_{1,2g-1} + s_{1,2g-1} + gk_{2g}+ gk_{2g+1} +  s_{2,2g-1} + k_{2g}s_{1,2g-1} + k_{2g+1}s_{1,2g}\\
&=\frac{1}{2}g(g+1) + gs_{1,2g+1} + s_{2,2g+1}
\end{align*}
where Lemma \ref{lem:sym} was used in the last step.

Let us now compute $v_3(K).$  We shall make use of the fact that $v_3$ satisfies the following skein relation \cite{IW,Lescop}:
\begin{equation}\label{eq:skein3}
v_3(K_+) - v_3(K_-) = \frac{1}{2}\Big(a_2(K_+)+a_2(K_-)+\mathrm{lk}^2(K',K'')\Big)-a_2(K')-a_2(K'')
\end{equation}
Again we proceed by induction, noting that for $g=0$, $v_3$ of the unknot is zero.  Suppose $g \geq 1$.  As before, we use (\ref{eq:skein3}) on a crossing in the last strand and see:
\begin{multline}\nonumber
v_3(K(k_1,\dots,k_{2g+1})) - v_3(K(k_1,\dots,k_{2g+1}-1)) = -\frac{1}{2}\Big(a_2(K(k_1,\dots,k_{2g+1})) \\ + a_2(K(k_1,\dots,k_{2g+1}-1))+(s_{1,2g}+g)^2\Big)
\end{multline}
Repeatedly applying (\ref{eq:skein3}) to the rest of the crossings in the last strand, one finds that:
\begin{multline}\nonumber
v_3(K(k_1,\dots,k_{2g+1})) - v_3(K(k_1,\dots,k_{2g},0)) = \\-\frac{1}{2}\Big(a_2(K(k_1,\dots,k_{2g+1})) + a_2(K(k_1,\dots,k_{2g},0)) \\+ 2\sum_{j=1}^{k_{2g+1}-1}a_2(K(k_1,\dots,k_{2g},j))+k_{2g+1}(s_{1,2g}+g)^2\Big)
\end{multline}
Applying the formula we have obtained for $a_2$ together with Lemma \ref{lem:sym} gives:
\begin{multline}\nonumber
v_3(K(k_1,\dots,k_{2g+1})) - v_3(K(k_1,\dots,k_{2g},0)) = \\-\frac{1}{2}\bigg(\frac{1}{2}g(g+1) + g(s_{1,2g}+k_{2g+1}) + s_{2,2g}+k_{2g+1}s_{1,2g} + \frac{1}{2}g(g+1) + gs_{1,2g} + s_{2,2g} \\+ 2\sum_{j=1}^{k_{2g+1}-1}\left(\frac{1}{2}g(g+1) + g(s_{1,2g}+j) + s_{2,2g}+js_{1,2g}\right)+k_{2g+1}(s_{1,2g}+g)^2\bigg) \\
=-\frac{1}{2}\Big(g(g+1) + 2gs_{1,2g}+ gk_{2g+1} + 2s_{2,2g}+k_{2g+1}s_{1,2g}\\+ (k_{2g+1}-1)g(g+1) + 2g(k_{2g+1}-1)s_{1,2g}+2(k_{2g+1}-1)s_{2,2g} \\+gk_{2g+1}(k_{2g+1}-1) + k_{2g+1}(k_{2g+1}-1)s_{1,2g}+k_{2g+1}(s_{1,2g}+g)^2\Big)\\
=-\frac{1}{2}\Big(k_{2g+1}g(g+1) + 2gk_{2g+1}s_{1,2g}+ gk_{2g+1}^2  + 2k_{2g+1}s_{2,2g} \\+k_{2g+1}^2s_{1,2g} + k_{2g+1}\left(s_{1,2g}^2+2gs_{1,2g}+g^2\right)\Big)\\
=-\frac{1}{2}k_{2g+1}\Big(g(2g+1) + 4gs_{1,2g}+ gk_{2g+1} + 2s_{2,2g} +k_{2g+1}s_{1,2g} + s_{1,2g}^2\Big)
\end{multline}
Similarly, applying \eqref{eq:skein3} repeatedly to the $2g$th strand now gives:
\begin{multline}\label{eq:ind3}
v_3(K(k_1,\dots,k_{2g+1})) - v_3(K(k_1,\dots,k_{2g-1},0,0)) = \\
-\frac{1}{2}k_{2g}\Big(g(2g+1) + 4gs_{1,2g-1}+ gk_{2g} + 2s_{2,2g-1} +k_{2g}s_{1,2g-1} + s_{1,2g-1}^2\Big)\\
-\frac{1}{2}k_{2g+1}\Big(g(2g+1) + 4gs_{1,2g}+ gk_{2g+1} + 2s_{2,2g} +k_{2g+1}s_{1,2g} + s_{1,2g}^2\Big)
\end{multline}
Now, applying \eqref{eq:skein3} to one of the two rightmost crossings of $K(k_1,\dots,k_{2g-1},0,0)$, one sees that:
\begin{multline}\nonumber
v_3(K(k_1,\dots,k_{2g-1},0,0)) - v_3(K(k_1,\dots,k_{2g-1})) = \\
-\frac{1}{2}\left(a_2(K(k_1,\dots,k_{2g-1},0,0))+a_2(K(k_1,\dots,k_{2g-1})+(s_{1,2g-1}+g)^2\right) \\
=-\frac{1}{2}\bigg(\frac{g(g+1)}{2} + gs_{1,2g-1} + s_{2,2g-1}+\frac{g(g-1)}{2} + (g-1)s_{1,2g-1} \\+ s_{2,2g-1}+s_{1,2g-1}^2 + 2gs_{1,2g-1}+g^2\bigg)\\
=-\frac{1}{2}\left(2g^2 + (4g-1)s_{1,2g-1} + 2s_{2,2g-1} + s_{1,2g}^2\right)\\
\end{multline}
Combining this with \eqref{eq:ind3} and applying the induction hypothesis gives:
\begin{multline}\nonumber
v_3(K(k_1,\dots,k_{2g+1})) =  v_3(K(k_1,\dots,k_{2g-1})) -\frac{1}{2}\left(2g^2 + (4g-1)s_{1,2g-1} + 2s_{2,2g-1} + s_{1,2g-1}^2\right) \\
-\frac{1}{2}k_{2g}\left(g(2g+1) + 4gs_{1,2g-1}+ gk_{2g} + 2s_{2,2g-1} +k_{2g}s_{1,2g-1} + s_{1,2g-1}^2\right)\\
-\frac{1}{2}k_{2g+1}\left(g(2g+1) + 4gs_{1,2g}+ gk_{2g+1} + 2s_{2,2g} +k_{2g+1}s_{1,2g} + s_{1,2g}^2\right)\\
=-\frac{1}{2}\bigg(\frac{g(g-1)(2g-1)}{3} + (g-1)(2g-1)s_{1,2g-1} + (g-1) s_{1,2g-1}^2 + 2(g-1) s_{2,2g-1} + s_{1,2g-1}s_{2,2g-1} + s_{3,2g-1}\\+2g^2 + (4g-1)s_{1,2g-1} + 2s_{2,2g-1} + s_{1,2g-1}^2\\+ k_{2g}\left(g(2g+1) + 4gs_{1,2g-1}+ gk_{2g} + 2s_{2,2g-1} +k_{2g}s_{1,2g-1} + s_{1,2g-1}^2\right) \\+k_{2g+1}\left(g(2g+1) + 4gs_{1,2g}+ gk_{2g+1} + 2s_{2,2g} +k_{2g+1}s_{1,2g} + s_{1,2g}^2\right)\bigg)\\
=-\frac{1}{2}\bigg(\frac{g(g+1)(2g+1)}{3} + g(2g+1)s_{1,2g-1} + g s_{1,2g-1}^2 + 2g s_{2,2g-1} + s_{1,2g-1}s_{2,2g-1} + s_{3,2g-1}\\
+ g(2g+1)k_{2g} + 4gk_{2g}s_{1,2g-1}+ gk_{2g}^2+ 2k_{2g}s_{2,2g-1} +k_{2g}^2s_{1,2g-1} + k_{2g}s_{1,2g-1}^2 \\
+g(2g+1)k_{2g+1} + 4gk_{2g+1}s_{1,2g}+ gk_{2g+1}^2 + 2k_{2g+1}s_{2,2g} +k_{2g+1}^2s_{1,2g} + k_{2g+1}s_{1,2g}^2\bigg)\\
=-\frac{1}{2}\bigg(\frac{g(g+1)(2g+1)}{3} + g(2g+1)(s_{1,2g-1}+ k_{2g} + k_{2g+1}) \\+ g \left(s_{1,2g-1}^2 + 2k_{2g}s_{1,2g-1} +k_{2g}^2 + 2k_{2g+1}s_{1,2g} +  k_{2g+1}^2\right)\\ + 2g( s_{2,2g-1} + k_{2g}s_{1,2g-1} + k_{2g+1}s_{1,2g}) + s_{3,2g-1}
 + k_{2g}s_{2,2g-1} + k_{2g+1}s_{2,2g} \\
+ s_{1,2g-1}s_{2,2g-1}+k_{2g}s_{1,2g-1}(s_{1,2g-1} + k_{2g}) + k_{2g}s_{2,2g-1}  + k_{2g+1}s_{2,2g} +k_{2g+1}s_{1,2g}(s_{1,2g} + k_{2g+1})\bigg)
\end{multline}
We now repeatedly apply Lemma \ref{lem:sym} to find:
\begin{multline}\nonumber
v_3(K(k_1,\dots,k_{2g+1})) = -\frac{1}{2}\bigg(\frac{g(g+1)(2g+1)}{3} + g(2g+1)s_{1,2g+1} \\+ g \left((s_{1,2g-1}+k_{2g})^2 + 2k_{2g+1}s_{1,2g} +  k_{2g+1}^2\right) + 2g( s_{2,2g}+ k_{2g+1}s_{1,2g}) + s_{3,2g}+ k_{2g+1}s_{2,2g} \\
+ s_{1,2g-1}s_{2,2g-1}+ k_{2g}s_{2,2g-1}+k_{2g}s_{1,2g-1}s_{1,2g}   + k_{2g+1}s_{2,2g} +k_{2g+1}s_{1,2g}s_{1,2g+1}\bigg)\\
= -\frac{1}{2}\bigg(\frac{g(g+1)(2g+1)}{3} + g(2g+1)s_{1,2g+1} + g \left(s_{1,2g}^2 + 2k_{2g+1}s_{1,2g} +  k_{2g+1}^2\right) + 2gs_{2,2g+1} + s_{3,2g+1} \\
+ s_{2,2g-1}(s_{1,2g-1}+ k_{2g})+k_{2g}s_{1,2g-1}s_{1,2g}   + k_{2g+1}s_{2,2g} +k_{2g+1}s_{1,2g}s_{1,2g+1}\bigg)\\
= -\frac{1}{2}\bigg(\frac{g(g+1)(2g+1)}{3} + g(2g+1)s_{1,2g+1} + g (s_{1,2g} + k_{2g+1})^2 + 2gs_{2,2g+1} + s_{3,2g+1} \\
+ s_{2,2g-1}s_{1,2g}+k_{2g}s_{1,2g-1}s_{1,2g}   + k_{2g+1}s_{2,2g} +k_{2g+1}s_{1,2g}s_{1,2g+1}\bigg)\\
= -\frac{1}{2}\bigg(\frac{g(g+1)(2g+1)}{3} + g(2g+1)s_{1,2g+1} + gs_{1,2g+1}^2 + 2gs_{2,2g+1} + s_{3,2g+1} \\
+ s_{1,2g}(s_{2,2g-1}+k_{2g}s_{1,2g-1})   + k_{2g+1}s_{2,2g} +k_{2g+1}s_{1,2g}s_{1,2g+1}\bigg)\\
= -\frac{1}{2}\bigg(\frac{g(g+1)(2g+1)}{3} + g(2g+1)s_{1,2g+1} + gs_{1,2g+1}^2 + 2gs_{2,2g+1} + s_{3,2g+1} \\
+ s_{1,2g}s_{2,2g} + k_{2g+1}s_{2,2g} +k_{2g+1}s_{1,2g}s_{1,2g+1}\bigg)\\
= -\frac{1}{2}\bigg(\frac{g(g+1)(2g+1)}{3} + g(2g+1)s_{1,2g+1} + gs_{1,2g+1}^2 + 2gs_{2,2g+1} + s_{3,2g+1} \\
+ s_{2,2g}(s_{1,2g} + k_{2g+1}) +k_{2g+1}s_{1,2g}s_{1,2g+1}\bigg)\\
= -\frac{1}{2}\bigg(\frac{g(g+1)(2g+1)}{3} + g(2g+1)s_{1,2g+1} + gs_{1,2g+1}^2 + 2gs_{2,2g+1} + s_{3,2g+1} \\
+ s_{2,2g}s_{1,2g+1} +k_{2g+1}s_{1,2g}s_{1,2g+1}\bigg)\\
= -\frac{1}{2}\bigg(\frac{g(g+1)(2g+1)}{3} + g(2g+1)s_{1,2g+1} + gs_{1,2g+1}^2 + 2gs_{2,2g+1} + s_{3,2g+1} \\
+ s_{1,2g+1}(s_{2,2g} +k_{2g+1}s_{1,2g})\bigg)\\
= -\frac{1}{2}\left(\frac{g(g+1)(2g+1)}{3} + g(2g+1)s_{1,2g+1} + gs_{1,2g+1}^2 + 2gs_{2,2g+1} + s_{3,2g+1}
+ s_{1,2g+1}s_{2,2g+1}\right)
\end{multline}
as desired.
\end{proof}

\subsection{Cosmetic Surgery Constraints}
We now review some of the main constraints on the existence of chirally cosmetic surgeries involving finite type invariants that are known.
Studying the degree 2 part of the LMO invariant, Ito has obtained the following:
\begin{thm}[Corollary 1.3 of \cite{Ito}]\label{thm:ft}
Let $K$ be a knot and suppose $S^3_{p/q}(K) \cong \pm S^3_{p/q'}(K).$  If $v_3(K) \neq 0,$ then
\[
\frac{p}{q+q'} = \frac{7a_2(K)^2 - a_2(K)-10a_4(K)}{2v_3(K)}
\]
\end{thm}
\begin{proof}[Remark.]\let\qed\relax
Ito uses a slightly different definition for $v_3$ than is used here; in particular, he normalizes it to take the value 1/4 on the right-handed trefoil. This means that ours differs from his by a factor of 4, and in fact, the statement in \cite{Ito} has $8v_3$ in the denominator of the right-hand-side instead.
\end{proof}

On the other hand, by combining Casson-Gordon and Casson-Walker invariants, Ichihara, Ito, and Saito find:
\begin{thm}[Theorem 1.2 of \cite{IIS}]\label{thm:cass}
Let $K$ be a knot and suppose $S^3_{p/q}(K) \cong - S^3_{p/q'}(K).$ Then
\[
4(q+q')a_2(K)=-\sigma(K,p)
\]
\end{thm}

Here $\sigma(K,p)$ is the \textit{p-signature} of the knot $K$.  We briefly recall the definition and basic properties of the signature.  If $A$ is any matrix that represents the Seifert form of $K$, then for each $\omega \in S^1 \subseteq \mathbf{C}$ on the complex unit circle, we put $\sigma_{\omega}(K)$ equal to the signature of the Hermitian matrix $(1-\omega)A+(1-\overline{\omega})A^T$.  This is the so-called \textit{signature function} of $K$ on the unit circle.  It is integer-valued, and it only has discontinuities at roots of the Alexander polynomial, which can be seen by rewriting the matrix in the definition as $(\overline{\omega}-1)\left(\omega A-A^T\right)$.  Near $\omega = 1$, this function is zero, and near $\omega = -1$, this function is equal to and invariant called simply the \textit{signature} of the knot.  In our case, for $K=K(k_1,\dots,k_{2g+1})$ it is known that the signature is equal to $2g$.  As the ``jump" in the signature function is equal to two at a simple root of the Alexander polynomial, whenever such a knot has Alexander polynomial with no repeated roots (as will be the case for the relevant knots below), the intervals between the roots will attain all nonnegative even integers up to $2g$ as signatures; see Figure \ref{fig:sig} for an illustration of the case $K=K(1,0,0,0,0)$.  Finally, the \textit{p-signature} of a knot, denoted $\sigma(K,p)$ is equal to the sum of the values signature function at the $p$th roots of unity.  That is,
\[
\sigma(K,p) = \sum_{\omega^p=1} \sigma_{\omega}(K)
\]

\begin{figure}
\centering
\includegraphics[width=.5\textwidth]{Signature_func}
\caption{The signature function for the knot $K(1,0,0,0,0)$.  The marked points are roots of the Alexander polynomial.}
\label{fig:sig}
\end{figure}

Following \cite{IIS}, we combine Theorems \ref{thm:ft} and \ref{thm:cass} to find that, if a knot $K$ admits a pair of chirally cosmetic surgeries along slopes $p/q$ and $p/q'$, then
\[
-\frac{\sigma(K,p)}{4a_2(K)} = q+q'=\frac{2pv_3(K)}{7a_2(K)^2 - a_2(K)-10a_4(K)}
\]
So we have obtained the following useful criterion:
\begin{cor}\label{cor:strong}
Let $K$ be a knot and suppose $S^3_{p/q}(K) \cong - S^3_{p/q'}(K).$ Then
\[
\frac{\sigma(K,p)}{p} = \frac{-8a_2(K)v_3(K)}{7a_2(K)^2 - a_2(K)-10a_4(K)}
\]
\end{cor}

\section{Proof of Theorem \ref{thm:main}}
We now prove our main result:

\main*
\noindent
We treat the five-stranded (genus 2) case first:
\subsection{The genus 2 case: $K=K(k_1,k_2,k_3,k_4,k_5)$}
In this case, by \eqref{eq:a242} and Lemma \ref{lem:av}

\begin{equation}\label{eq:ft2}
\begin{cases}
a_2(K) = 3 + 2 s_{1,5} + s_{2,5}\\
a_4(K) = 1 + s_{1,5} + s_{2,5} + s_{3,5} + s_{4,5}\\
v_3(K) = -\frac{1}{2}\left(10+10s_{1,5}+2s_{1,5}^2+4s_{2,5}+s_{1,5}s_{2,5}+s_{3,5}\right)
\end{cases}
\end{equation}

We see that $a_2(K),a_4(K)>0$ and $v_3(K)<0.$  Moreover, we also have that $0<\sigma(K,p)\leq 4p$ for all $p>0$.  So if $K$ admits chirally cosmetic surgeries with slopes $p/q$ and $p/q',$ then, by Corollary \ref{cor:strong},
\begin{align*}
0 < \frac{\sigma(K,p)}{p} &= \frac{-8a_2(K)v_3(K)}{7a_2(K)^2 - a_2(K)-10a_4(K)} \leq 4 \\
\leadsto 2a_2(K)|v_3(K)| & \leq 7a_2(K)^2 - a_2(K)-10a_4(K)\numberthis \label{eq:ineq}
\end{align*}
We also have, by \eqref{eq:ft2} the following estimate relating $a_2(K)$ and $a_4(K)$:
\[
a_4(K) = 1 + s_{1,5} + s_{2,5} + s_{3,5} + s_{4,5} \geq 1+ \frac{2}{3}s_{1,5} + \frac{1}{3}s_{2,5} = \frac{1}{3}a_2(K)
\]
Hence, if $K$ admits any chirally cosmetic surgery, then:
\begin{align*}
2a_2(K)|v_3(K)| & \leq 7a_2(K)^2 - a_2(K)-10a_4(K) < 7a_2(K)^2 - a_2(K)-3a_2(K) \\
\leadsto 2|v_3(K)| &< 7a_2(K) - 4
\end{align*}
Thus, we have shown:
\begin{cor}\label{cor:weak}
Let $K=K(k_1,k_2,k_3,k_4,k_5)$.  If
\[
2|v_3(K)| \geq 7a_2(K) - 4
\]
then $K$ does not admit any chirally cosmetic surgeries.
\end{cor}
We now apply this corollary to exclude all but four knots in this family from admitting chirally cosmetic surgeries.  We proceed in three cases:\\
\textbf{Case I: At least three $k_i$ are nonzero:}
By \eqref{eq:ft2} we see that
\begin{align*}
2|v_3(K)| - 7a_2(K) + 4 &= 10 +	10 s_1 + 2 s_1^2 +4 s_2 +s_1 s_2 +s_3 - 21 -14 s_1 -7 s_2 +4\\
&=-7-4s_1+2s_1^2-3s_2+s_1s_2+s_3\\
&=-7+(2s_1-4)s_1+(s_1-3)s_2 + s_3
\end{align*}

In this case, we have that $s_1 \geq 3$ and $s_3 \geq 1$ so that
\[
2|v_3(K)| - 7a_2(K) + 4 \geq -7+(6-4)3 + 1 =0
\]
Hence, by Corollary \ref{cor:weak}, these knots admit no chirally cosmetic surgeries.\\
\textbf{Case II: Exactly two $k_i$ are nonzero:}
Notice that, without loss of generality, we may assume $(k_1,k_2,k_3,k_4,k_5) = (a,b,0,0,0)$ for $a \geq b >0$.  This is because, by the symmetries of pretzel knots, one may cyclically permute the strands until the longest strand is in the first position; then, by applying ``flyping" moves, one can bring the other ``long" strand to the second position.  We compute that $s_1 = a+b,$ $s_2=ab,$ and $s_3=0.$

Suppose $a+b\geq 4.$ Then
\begin{align*}
|v_3(K)| &= 5+5(a+b)+(a+b)^2 + 2ab+ \frac{1}{2}(a+b)(ab) \\
& \geq 5 + 9(a+b) + \frac{ab}{2}(4+a+b)
\end{align*}
Thus
\begin{align*}
2|v_3(K)| - 7a_2(K) &\geq 10 + 18(a+b) + ab(4+a+b) - 21 -14 (a+b) -7 ab +4\\
&=-7+4(a+b)+ab(a+b-3)\\
&geq -7+16 + ab \geq 0
\end{align*}
Once again, Corollary \ref{cor:weak} guarantees that $K(a,b,0,0,0)$ admits no chirally cosmetic surgeries as long as $a+b \geq 4.$  This leaves: $K(1,1,0,0,0)$ and $K(2,1,0,0,0)$ as still unchecked; we deal with these two knots below.\\
\textbf{Case III: Exactly one $k_i$ nonzero:}
As before, we may assume (by appropriate permutation) that the longest strand is in the first position; i.e., we are considering the case $K(N,0,0,0,0)$ with $N > 0.$  Here $s_1=N$ and $s_2=s_3=s_4=0.$  Thus, by \eqref{eq:ft2}:
\begin{align*}
7a_2(K)^2 - a_2(K) - 10a_4(K) &= 28N^2+72N+50 \\
2a_2(K)|v_3(K)| &= 4N^3+26N^2+50N+30
\end{align*}
We see that \eqref{eq:ineq} is satisfied only for $N \leq 2,$ and so, by Corollary \ref{cor:strong}, $K(N,0,0,0,0)$ admits no chirally cosmetic sugeries when $N \geq 3.$  This leaves $K(1,0,0,0,0)$ and $K(2,0,0,0,0).$  As still unchecked.  We turn to these and the other two leftover knots now.\\
\textbf{The remaining knots:}
Let us turn our attention to the four knots that still need to be checked.

First, we consider $K=K(2,1,0,0,0).$  We compute, using \eqref{eq:ft2} that
\begin{align*}
2a_2(K)|v_3(K)| &= 2 (11)(36) = 792\\
7a_2(K)^2 - a_2(K) - 10a_4(K) &= 7(121) -11-60 = 776
\end{align*}
The inequality \eqref{eq:ineq} is not satisfied, and so, by Corollary \ref{cor:strong}, this knot does not admit any chirally cosmetic surgeries.

For the next three knots, we will investigate their signature functions in order to obtain bounds for their respective p-signatures.

Let $K=K(1,0,0,0,0).$  We compute that $a_2(K)=5$, $a_4(K)=2$, and $v_3(K)=-11$
\[
\frac{-8a_2(K)v_3(K)}{7a_2(K)^2 - a_2(K)-10a_4(K)} = \frac{44}{15}.
\]
Hence, by Corollary \ref{cor:strong}, if $K$ admits chirally cosmetic surgeries with slopes $p/q$ and $p/q',$ then $p=15n$ for some positive integer $n$.  From the Conway polynomial $1+5z^2+2z^4,$ we find that the roots of the Alexander polynomial are $e^{\pm i\theta_1}$ and $e^{\pm i\theta_2},$ where $\theta_1 \approx 1.712>\frac{8\pi}{15}$ and $\theta_2 \approx 0.473>\frac{2\pi}{15}.$  Hence, $\sigma_\omega(K)=0$ when $-\frac{2\pi}{15}\leq \arg(\omega)\leq \frac{2\pi}{15}$ and $\sigma_\omega(K)\leq 2$ when $-\frac{8\pi}{15}\leq \arg(\omega)\leq \frac{8\pi}{15}.$  See Figure \ref{fig:sig} for an illustration.

So
\begin{align*}
\frac{\sigma(K,15n)}{15n} &\leq \frac{2(6n)+4(7n-1)}{15n}\\
&=\frac{40n-4}{15n}\\
&<\frac{40}{15} < \frac{44}{15}
\end{align*}
Hence, by Corollary \ref{cor:strong}, $K$ admits no chirally cosmetic surgeries.

Now let $K=K(2,0,0,0,0).$  We compute $a_2(K)=7$, $a_4(K)=3$, and $v_3(K)=-19$
\[
\frac{-8a_2(K)v_3(K)}{7a_2(K)^2 - a_2(K)-10a_4(K)} = \frac{532}{153}.
\]
As before, by Corollary \ref{cor:strong}, if $K$ admits chirally cosmetic surgeries with slopes $p/q$ and $p/q',$ then $p=153n$ for some positive integer $n$.  By \eqref{eq:a242}, this knot has Conway polynomial $1+7z^2+3z^4$.  Thus we find that the roots of the Alexander polynomial, which we denote $e^{\pm i\theta_1}$ and $e^{\pm i\theta_2}$, satisfy $\theta_1 \approx 1.661>\frac{80\pi}{153}$ and $\theta_2 \approx 0.394>\frac{18\pi}{153}$.  Hence, $\sigma_\omega(K)=0$ when $-\frac{18\pi}{153}\leq \arg(\omega)\leq \frac{18\pi}{153}$ and $\sigma_\omega(K)\leq 2$ when $-\frac{80\pi}{153}\leq \arg(\omega)\leq \frac{80\pi}{153}$.
So
\begin{align*}
\frac{\sigma(K,153n)}{153n} &\leq \frac{2(62n)+4(73n-1)}{15n}\\
&=\frac{416n-4}{153n}\\
&<\frac{416}{153} < \frac{532}{153}
\end{align*}
Hence, by Corollary \ref{cor:strong}, $K$ admits no chirally cosmetic surgeries.

Lastly, let $K=K(1,1,0,0,0).$  We compute $a_2(K)=8$, $a_4(K)=4$, and $v_3(K)=-22$
\[
\frac{-8a_2(K)v_3(K)}{7a_2(K)^2 - a_2(K)-10a_4(K)} = \frac{88}{25}.
\]
As before, by Corollary \ref{cor:strong}, if $K$ admits chirally cosmetic surgeries with slopes $p/q$ and $p/q',$ then $p=25n$ for some positive integer $n$.  From the Conway polynomial $1+8z^2+4z^4,$ we find that the roots of the Alexander polynomial are $e^{\pm i\theta_1}$ and $e^{\pm i\theta_2},$ where $\theta_1 \approx 1.504>\frac{10\pi}{25}$ and $\theta_2 \approx 0.368>\frac{2\pi}{25}.$  Hence, $\sigma_\omega(K)=0$ when $-\frac{2\pi}{25}\leq \arg(\omega)\leq \frac{2\pi}{25}$ and $\sigma_\omega(K)\leq 2$ when $-\frac{10\pi}{25}\leq \arg(\omega)\leq \frac{10\pi}{25}.$
So
\begin{align*}
\frac{\sigma(K,25n)}{25n} &\leq \frac{2(8n)+4(15n-1)}{15n}\\
&=\frac{76n-4}{25n}\\
&<\frac{76}{25} < \frac{88}{25}
\end{align*}
Once again, by Corollary \ref{cor:strong}, $K$ admits no chirally cosmetic surgeries.  This concludes the case of genus 2.

\subsection{The genus 3 case: $K=K(k_1,k_2,k_3,k_4,k_5,k_6,k_7)$}
In this case, by \eqref{eq:a243}, and Lemma \ref{lem:av}

\begin{equation}\label{eq:ft3}
\begin{cases}
a_2(K) = 6 + 3 s_{1,7} + s_{2,7}\\
a_4(K) = 5 + 4s_{1,7} + 3s_{2,7} + 2s_{3,7} + s_{4,7}\\
a_6(K) = 1 + s_{1,7} + s_{2,7} + s_{3,7} + s_{4,7} + s_{5,7} + s_{6,7}\\
v_3(K) = -\frac{1}{2}\left(28+21s_{1,5}+3s_{1,5}^2+6s_{2,5}+s_{1,5}s_{2,5}+s_{3,5}\right)
\end{cases}
\end{equation}

Once again $a_2(K),a_4(K)>0$ and $v_3(K)<0.$  Moreover, we also have that $0<\sigma(K,p)\leq 6p$ for all $p>0$.  So if $K$ admits chirally cosmetic surgeries with slopes $p/q$ and $p/q',$ then, by Corollary \ref{cor:strong},
\begin{align*}
0 < \frac{\sigma(K,p)}{p} &= \frac{-8a_2(K)v_3(K)}{7a_2(K)^2 - a_2(K)-10a_4(K)} \leq 6 \\
\leadsto \frac{4}{3}a_2(K)|v_3(K)| & \leq 7a_2(K)^2 - a_2(K)-10a_4(K)\numberthis \label{eq:ineq3}
\end{align*}
We also have, by \eqref{eq:ft3} the following estimate relating $a_2(K)$ and $a_4(K)$:
\[
\frac{6}{5}a_4(K) = 6 + \frac{24}{5}s_{1,7} + \frac{18}{5}s_{2,7} + \frac{12}{5}s_{3,7} + \frac{6}{5}s_{4,7} \geq 6+ 3s_{1,7} + s_{2,7} = a_2(K)
\]
Hence, if $K$ admits any chirally cosmetic surgery, then:
\begin{align*}
\frac{4}{3}a_2(K)|v_3(K)| & \leq 7a_2(K)^2 - a_2(K)-10a_4(K) \leq 7a_2(K)^2 - a_2(K)-\frac{25}{3}a_2(K) \\
\leadsto 4|v_3(K)| & \leq  21a_2(K) - 28
\end{align*}
Thus, we have shown:
\begin{cor}\label{cor:weak3}
Let $K=K(k_1,k_2,k_3,k_4,k_5,k_6,k_7)$.  If
\[
4|v_3(K)| > 21a_2(K) - 28
\]
then $K$ does not admit any chirally cosmetic surgeries.
\end{cor}
Now we compute:
\begin{align*}
4|v_3(K)| - 21a_2(K)+ 28 &=-42-21s_{1,7}-9s_{2,7}+6s_{1,7}^2+2s_{1,7}s_{2,7}+2s_{3,7}\\
&=-42 + 3s_{1,7}(2s_{1,7}-7) + s_{2,7}(2s_{1,7}-9)+2s_{3,7}
\end{align*}
If $s_{1,7} \geq 5$, then
\[
4|v_3(K)| - 21a_2(K)+ 28 \geq -42+3(5)(3) > 0
\]
Hence, by Corollary \ref{cor:weak3}, these knots do not admit any chirally cosmetic surgeries.  It remains to check the knots in this family with $s_{1,7}<5$.  These are (once again using the symmetries of the pretzel knots): $K(1,1,1,1,0,0,0)$, $K(2,1,1,0,0,0,0)$, $K(1,1,1,0,0,0,0)$, $K(3,1,0,0,0,0,0)$, $K(2,2,0,0,0,0,0)$, $K(2,1,0,0,0,0,0)$, $K(1,1,0,0,0,0,0)$, $K(4,0,0,0,0,0,0)$, $K(3,0,0,0,0,0,0)$, $K(2,0,0,0,0,0,0)$, and $K(1,0,0,0,0,0,0)$.

First, we note that for $K=K(1,1,1,1,0,0,0)$, we have, by \eqref{eq:ft3}, $a_2(K)=24$, $a_4(K)=48$, and $v_3(K)=-112$.  Thus, $\frac{4}{3}a_2(K)|v_3(K)| = 3584 > 3528 = 7a_2(K)^2 - a_2(K)-10a_4(K)$ so that inequality \ref{eq:ineq3} is not satisfied, implying that this knot admits no chirally cosmetic surgeries.

For the rest of the knots, as in the genus 2 case, we compute $F \coloneqq \frac{-8a_2(K)v_3(K)}{7a_2(K)^2 - a_2(K)-10a_4(K)}$.  By Corollary \ref{cor:strong}, if $K$ admits chirally cosmetic surgeries with slopes $p/q$ and $p/q'$, then $\frac{\sigma(K,p)}{p}=F$ so that $p$ must be a multiple of the denominator of $F$.   From the roots of their Alexander polynomials, we obtain bounds on their signature functions. For all but one of the remaining knots, we find that $\frac{\sigma(K,p)}{p}<F$ so that those knots admit no chirally cosmetic surgeries.  The results are summarized in the following table (we denote the roots of the Alexander polynomial by $e^{\pm i\theta_1}$,  $e^{\pm i\theta_2}$, and  $e^{\pm i\theta_3}$):\\
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$K$ & $F$ & $\theta_1 >$ & $\theta_2 >$ & $\theta_3 >$ & $\frac{\sigma(K,p)}{p}\leq$\\
\hline
$K(2,1,1,0,0,0,0)$ & $\frac{1219}{205}$ & $\frac{14\pi}{205}$ & $\frac{54\pi}{205}$ & $\frac{122\pi}{205}$ & $ \frac{6(83n-1)+4(68n)+2(40n)}{205n}<\frac{850}{205}$\\
\hline
$K(1,1,1,0,0,0,0)$ & $\frac{5256}{985}$ & $\frac{76\pi}{985}$ & $\frac{282\pi}{985}$ & $\frac{600\pi}{985}$ & $ \frac{6(385n-1)+4(318n)+2(206n)}{985n}<\frac{3994}{985}$\\
\hline
$K(3,1,0,0,0,0,0)$ & $\frac{2660}{461}$ & $\frac{32\pi}{461}$ & $\frac{140\pi}{461}$ & $\frac{294\pi}{461}$ & $ \frac{6(167n-1)+4(154n)+2(108n)}{461n}<\frac{1834}{461}$\\
\hline
$K(2,2,0,0,0,0,0)$ & $\frac{400}{69}$ & $\frac{4\pi}{69}$ & $\frac{20\pi}{69}$ & $\frac{42\pi}{69}$ & $ \frac{6(27n-1)+4(22n)+2(16n)}{69n}<\frac{282}{69}$\\
\hline
$K(2,1,0,0,0,0,0)$ & $\frac{578}{111}$ & $\frac{8\pi}{111}$ & $\frac{34\pi}{111}$ & $\frac{70\pi}{111}$ & $ \frac{6(41n-1)+4(36n)+2(26n)}{111n}<\frac{442}{111}$\\
\hline
$K(1,1,0,0,0,0,0)$ & $\frac{468}{101}$ & $\frac{8\pi}{101}$ & $\frac{32\pi}{101}$ & $\frac{64\pi}{101}$ & $ \frac{6(37n-1)+4(32n)+2(24n)}{101n}<\frac{398}{101}$\\
\hline
$K(4,0,0,0,0,0,0)$ & $\frac{96}{17}$ & $0$ & $\frac{4\pi}{17}$ & $\frac{10\pi}{17}$ & $ \frac{6(7n-1)+4(6n)+2(4n)}{17n}<\frac{74}{17}$\\
\hline
$K(3,0,0,0,0,0,0)$ & $\frac{708}{139}$ & $\frac{10\pi}{139}$ & $\frac{48\pi}{139}$ & $\frac{92\pi}{139}$ & $ \frac{6(47n-1)+4(44n)+2(38n)}{139n}<\frac{534}{139}$\\
\hline
$K(2,0,0,0,0,0,0)$ & $\frac{1968}{433}$ & $\frac{42\pi}{433}$ & $\frac{158\pi}{433}$ & $\frac{292\pi}{433}$ & $ \frac{6(141n-1)+4(134n)+2(116n)}{433n}<\frac{1614}{433}$\\
\hline
\end{tabular}
\end{center}
\hfill \\

Finally, we turn our attention to the knot $K(1,0,0,0,0,0,0)$.  By \eqref{eq:ft3}, we compute that: $\frac{-8a_2(K)v_3(K)}{7a_2(K)^2 - a_2(K)-10a_4(K)} = 4$.  The roots of the Alexander polynomial for this knot are $e^{\pm 2\pi i\alpha}$,  $e^{\pm 2\pi i\beta}$, and  $e^{\pm 2\pi i\gamma}$, where $.056 < \alpha < .057$, $.19 < \beta < .191$, and $.342 < \gamma < .343$.  Therefore, we obtain the estimate:
\begin{align*}
\frac{\sigma(K,p)}{p} &\leq \frac{6\lceil .316 p\rceil + 4(2\lceil.153p\rceil) + 2(2\lceil.135 p\rceil)}{p}\\
&\leq \frac{6(.316 p +1) + 4(2(.153p+1)) + 2(2(.135 p +1))}{p}\\
&= 3.66+\frac{18}{p}
\end{align*}
Hence, if $p\geq 53$, then $\frac{\sigma(K,p)}{p}<4$, which by Corollary \ref{cor:strong} excludes the possibility of chirally cosmetic surgeries on this knot with slopes $p/q$ and $p/q'$ whenever $p \geq 53$.  Using Mathematica \cite{WM}, we  explicitly compute $\frac{\sigma(K,p)}{p}$ for $1 \leq p \leq 52$. The results are in the table below:
\begin{center}
\begin{tabular}{|c|c||c|c||c|c||c|c|}
\hline
$p$ & $\frac{\sigma(K,p)}{p}$ & $p$ & $\frac{\sigma(K,p)}{p}$ & $p$ & $\frac{\sigma(K,p)}{p}$ & $p$ & $\frac{\sigma(K,p)}{p}$\\
\hline
 1 & 0 & 14 & $\frac{27}{7}$ & 27 & $\frac{32}{9}$ &
   40 & $\frac{73}{20}$ \\ \hline
 2 & 3 & 15 & $\frac{56}{15}$ & 28 & $\frac{51}{14}$ &
   41 & $\frac{148}{41}$ \\ \hline
 3 & $\frac{8}{3}$ & 16 & $\frac{29}{8}$ &
   29 & $\frac{108}{29}$ & 42 & $\frac{25}{7}$ \\ \hline
 4 & $\frac{7}{2}$ & 17 & $\frac{64}{17}$ &
   30 & $\frac{11}{3}$ & 43 & $\frac{156}{43}$ \\ \hline
 5 & 4 & 18 & $\frac{31}{9}$ & 31 & $\frac{116}{31}$ &
   44 & $\frac{79}{22}$ \\ \hline
 6 & 3 & 19 & $\frac{68}{19}$ & 32 & $\frac{59}{16}$ &
   45 & $\frac{164}{45}$ \\ \hline
 7 & $\frac{24}{7}$ & 20 & $\frac{37}{10}$ &
   33 & $\frac{40}{11}$ & 46 & $\frac{85}{23}$ \\ \hline
 8 & $\frac{15}{4}$ & 21 & $\frac{24}{7}$ &
   34 & $\frac{63}{17}$ & 47 & $\frac{172}{47}$ \\ \hline
 9 & $\frac{32}{9}$ & 22 & $\frac{39}{11}$ &
   35 & $\frac{132}{35}$ & 48 & $\frac{29}{8}$ \\ \hline
 10 & $\frac{19}{5}$ & 23 & $\frac{84}{23}$ &
   36 & $\frac{65}{18}$ & 49 & $\frac{180}{49}$ \\ \hline
 11 & $\frac{40}{11}$ & 24 & $\frac{43}{12}$ &
   37 & $\frac{132}{37}$ & 50 & $\frac{91}{25}$ \\ \hline
 12 & $\frac{7}{2}$ & 25 & $\frac{92}{25}$ &
   38 & $\frac{67}{19}$ & 51 & $\frac{188}{51}$ \\ \hline
 13 & $\frac{48}{13}$ & 26 & $\frac{49}{13}$ &
   39 & $\frac{140}{39}$ & 52 & $\frac{97}{26}$ \\ \hline
\end{tabular}
\end{center}

\hfill \\
We see that $\frac{\sigma(K,p)}{p}=4$ only when $p=5$.  Hence, by Corollary \ref{cor:strong}, if $K$ admits chirally cosmetic surgeries, they must have surgery slopes $5/q$ and $5/q'$ for some integers $q$ and $q'$.  However, by Theorem \ref{thm:cass}, we must have:
\[
q+q' = \frac{-\sigma(K,5)}{4a_2(K)} = \frac{-20}{4\cdot 9} = -\frac{5}{9}
\]
which is impossible, as $q$ and $q'$ are supposed to be integers.
\qed

\section{Proof of Theorem \ref{thm:gen}}

Now we turn our attention to the general case.  We show that for a fixed genus, at most finitely many alternating odd pretzel knots can possibly admit chirally cosmetic surgeries.  We shall make use of the following:
\begin{lem}[Corollary 6.2 of \cite{IIS}]\label{lem:weak}
Let $K$ be a nontrivial negative knot (i.e., a knot with all negative crossings).  If $4|v_3(K)| \geq 7g a_2(K)$ then $K$ admits no chirally cosmetic surgeries (here $g$ is the Seifert genus of $K$.
\end{lem}
\begin{proof}This follows from the fact that for negative knots, $a_2(K),a_4(K) \geq 0$ \cite{Crom}, $v_3(K)<0$ \cite{Stoi}, and $\sigma(K,p)>0$ \cite{PT}.  (For the family of knots we are considering, two of these inequalities can be seen directly from Lemma \ref{lem:av}).  Now Corollary \ref{cor:strong} implies that if $p/q$- and $p/q'$-surgeries are chirally cosmetic,
\begin{align*}
0 < \frac{\sigma(K,p)}{p} &= \frac{-8a_2(K)v_3(K)}{7a_2(K)^2 - a_2(K)-10a_4(K)} \leq 2g \\
\leadsto 4a_2(K)|v_3(K)| & \leq g(7a_2(K)^2 - a_2(K)-10a_4(K) < 7a_2(K)^2)\\
\leadsto 4|v_3(K)| &< 7ga _2(K)
\end{align*}
\end{proof}
We now prove:
\gen*
\begin{proof}
As these knots are negative, by Lemma \ref{lem:weak}, $K$ admits no chirally cosmetic surgeries if $4|v_3(K)|\geq 7g a_2(K)$ .  By Lemma \ref{lem:av}, we compute:
\begin{multline}\nonumber
4|v_3(K)|- 7g a_2(K) = \frac{2g(g+1)(2g+1)}{3}-\frac{7g(g+1)}{2} + 2g(2g+1)s_{1,2g+1} - 7g^2s_{1,2g+1} \\+ 4gs_{2,2g+1}-7gs_{2,2g+1} + 2gs_{1,2g+1}^2+2s_{1,2g+1}s_{2,2g+1} + 2s_{3,2g+1}\\
= \frac{-13g^3-9g^2+4g}{6} + g(2-3g)s_{1,2g+1}+ 2gs_{1,2g+1}^2 -3gs_{2,2g+1}+2s_{1,2g+1}s_{2,2g+1} + 2s_{3,2g+1}\\
=\frac{-13g^3-9g^2+4g}{6} + gs_{1,2g+1}(2s_{1,2g+1}+2-3g) + s_{2,2g+1}(2s_{1,2g+1}-3g) + 2s_{3,2g+1}
\end{multline}
If $s_{1,2g+1} \geq \alpha g$, then $s_{1,2g+1}(2s_{1,2g+1}-3g)\geq \frac{13g^2}{6}$, $2s_{1,2g+1}-3g \geq 0$, and $2gs_{1,2g+1} \geq 4g^2$. Hence:
\[
4|v_3(K)|- 7g a_2(K) \geq \frac{-13g^3-9g^2+4g}{6} + \frac{13g^3}{6}+4g^2 \geq 0
\]
as desired.
\end{proof}
\printbibliography

\end{document}

redpajama_set_name: RedPajamaArXiv


The Radisson Blu Hotel Pudong Century Park carried out 171 love boxes filled with goodies and candies to MiFan MaMa as the hotel guests and employees give their support to the "Boxes of Love 2012" charity campaign by MiFan MaMa.
With arrivals recording over a 800,000, the year 2011 was a resounding success for Sri Lanka Tourism's marketing campaign 'Visit Sri Lanka 2011'.
The Team of Tourism – Insider wishes all of you a Merry Christmas with your family and a Happy New Year!
We have also closed for Christmas and we will reopen again on the second of january 2012.
If in the meantime something important is happend, we will come back as soon as possible. Please check our page some times.
airberlin is further expanding the flights it offers to North America and will fly non-stop from Düsseldorf to Las Vegas twice a week as of summer 2012. With the new destination, airberlin passengers can now fly to a total of seven destinations in North America (New York, Miami, Fort Myers, Los Angeles, San Francisco, Las Vegas and Vancouver).
Etihad Airways, the national airline of the United Arab Emirates, has agreed to increase its stake in airberlin, Europe's sixth largest airline, to 29.21 per cent, becoming airberlin's largest single shareholder.


text:
Q: Как в discord.py сделать функцию которая будет выбирать один из двух вариантов? Пытаюсь сделать discord бота на python и хочу сделать функцию,которая будет рандомно выбирать либо "орёл" ,либо "решка",но не знаю как.

A: Можно использовать библиотеку random. Через нее уже куча вариантов решения, примеры:

Через выбор из списка:
import random
variants = ['орел', 'решка']
print(random.choice(variants))

Через случайное число:
Этот метод стоит использовать, когда вам нужно изменить шансы, то есть не только 50 на 50
import random
choice = random.randint(0, 100) # случайное число от 0 до 100
if choice < 30:
   print('Первый вариант (шанс выпадения 30%)')
else:
   print('Второй вариант (шанс выпадения 70%)')

Сама функция может быть оформлена примерно так:
import discord, random
from discord.ext import commands

bot = commands.Bot(command_prefix='!', intents = discord.Intents.all())

@bot.command()
async def орел_решка(ctx):
    await ctx.send(random.choice(['орел', 'решка']))

bot.run(TOKEN)

Ну или другими методами. Вызов функции в чате дискорда: !орел_решка
Не забудьте указать ваш токен в конце!

redpajama_set_name: RedPajamaStackExchange


text:
Felipe Ángeles är en ort i Mexiko.   Den ligger i kommunen Xicoténcatl och delstaten Tamaulipas, i den centrala delen av landet,  km norr om huvudstaden Mexico City. Felipe Ángeles ligger  meter över havet och antalet invånare är .

Terrängen runt Felipe Ángeles är platt, och sluttar söderut. Den högsta punkten i närheten är Sierra Cucharas,  meter över havet,  km väster om Felipe Ángeles. Runt Felipe Ángeles är det ganska glesbefolkat, med  invånare per kvadratkilometer. Närmaste större samhälle är Xicoténcatl,  km öster om Felipe Ángeles. Omgivningarna runt Felipe Ángeles är en mosaik av jordbruksmark och naturlig växtlighet.

Savannklimat råder i trakten. Årsmedeltemperaturen i trakten är  °C. Den varmaste månaden är maj, då medeltemperaturen är  °C, och den kallaste är januari, med  °C. Genomsnittlig årsnederbörd är  millimeter. Den regnigaste månaden är september, med i genomsnitt  mm nederbörd, och den torraste är januari, med  mm nederbörd.

Kommentarer

Källor

Orter i Tamaulipas

redpajama_set_name: RedPajamaWikipedia


Ikuko Maekawa moved from Nara city to Soni six years ago.She nurtured her dream of running her own cafe somedayfrom the time she was single. It was two years ago that sheopened the cafe.
It is located at the foot of Mt.Yoroi overlooking SoniKogen Highlands. "Cafe Nekoron" made by borrowing anold private house in this place and renovating the barninto a cafe. The origin of the shop name is combination ofcat that Ikuko likes and Macaroon that her husband likes.The reason she chose Soni Village is that it is close to herhusband's parents' house and also she thought "I want torun the fi rst cafe in the village".
It took another year from the renovation to the openingof the cafe. "Once the building was completed, I becameworried about such as the interior and the menu to serveat the shop. There isn't much point in opening a cafe in ahurry, so I performed tasks one by one over time.
The in-store menus are drinks such as coffee whichordered from Kanazawa and confectionery includinghandmade muffins baked by Ikuko. The books to whichIkuko has a strong attachment are lined up all over the cafe.Its choice is like a book cafe.
The music that fl ows in cafe is also pleasant to the earand this is "hideout" in Soni where people can enjoy slowand calm time.


text:
London's Markets - From Smithfield to Portobello Road
Stephen Halliday
Publisher: The History Press
London is a City of Markets: markets in meat, fish, fruit, vegetables, money, insurance, shipping and, occasionally, in stolen goods. Stephen Halliday's book is a comprehensive account of the often lurid and controversial history of its markets from Roman Londinium to the London of Boris Johnson as well as a guide to visiting them (and emerging with a bargain).
This is the page of Stephen Halliday on 24symbols. Here you can see and read his/her books.
Euripides was one of the greatest Greek tragedians and is considered one of the most important figures in ancient literature.  Euripides is thought to have written close to 100 plays and almost 20 of them have survived.  This edition of Heracles includes a table of contents.
The Great War - Indian Writings...
Rakhshanda Jalil
It is very hard to endure the bombs, Father. It will be difficult for anyone to survive and come back safe and sound from the war. The son who is very lucky will see his father and mother…
(Extract from a letter by an Indian soldier serving in France, written on 14 January 1915 to his father)
The Great War, as the First World War was referred to, saw the service of over 1.3 million Indians, of whom 74,000 never made it back home. For their families, the War was something they could not fully fathom. Soldiers from the Indian subcontinent won over 12,908 awards for bravery, including 11 Victoria Crosses. Yet this unprecedented show of valour by Indian soldiers remains largely unsung and unrecognised-particularly in India.
Commemorating hundred years of the end of the First World War, this volume brings together diverse voices-Rabindranath Tagore, Mulk Raj Anand, Sarojini Naidu, Mohamed Ali, Chandradhar Sharma Guleri and many more-that reflect a variety of attitudes among Indians towards the War. Included too are Rakhshanda Jalil's original translations of the works of Urdu poets of the time capturing their responses to the War. This volume of writings, originally written in Urdu, Hindi, Bengali and English, attempts to recognise and remember the contribution of the unknown soldiers to the Great War.
The Power of Cute
Simon May
An exploration of cuteness and its immense hold on us, from emojis and fluffy puppies to its more uncanny, subversive expressions
Cuteness has taken the planet by storm. Global sensations Hello Kitty and Pokémon, the works of artists Takashi Murakami and Jeff Koons, Heidi the cross-eyed opossum and E.T.—all reflect its gathering power. But what does "cute" mean, as a sensibility and style? Why is it so pervasive? Is it all infantile fluff, or is there something more uncanny and even menacing going on—in a lighthearted way? In The Power of Cute, Simon May provides nuanced and surprising answers.
We usually see the cute as merely diminutive, harmless, and helpless. May challenges this prevailing perspective, investigating everything from Mickey Mouse to Kim Jong-il to argue that cuteness is not restricted to such sweet qualities but also beguiles us by transforming or distorting them into something of playfully indeterminate power, gender, age, morality, and even species. May grapples with cuteness's dark and unpindownable side—unnerving, artful, knowing, apprehensive—elements that have fascinated since ancient times through mythical figures, especially hybrids like the hermaphrodite and the sphinx. He argues that cuteness is an addictive antidote to today's pressured expectations of knowing our purpose, being in charge, and appearing predictable, transparent, and sincere. Instead, it frivolously expresses the uncertainty that these norms deny: the ineliminable uncertainty of who we are; of how much we can control and know; of who, in our relations with others, really has power; indeed, of the very value and purpose of power.
The Power of Cute delves into a phenomenon that speaks with strange force to our age.
King Warrior Magician Lover -...
Robert Moore, Doug Gillette
The bestselling, widely heralded, Jungian introduction to the psychological foundation of a mature, authentic, and revitalized masculinity.
Redefining age-old concepts of masculinity, Jungian analysts Robert Moore and Douglas Gillette make the argument that mature masculinity is not abusive or domineering, but generative, creative, and empowering of the self and others. Moore and Gillette clearly define the four mature male archetypes that stand out through myth and literature across history: the king (the energy of just and creative ordering), the warrior (the energy of aggressive but nonviolent action), the magician (the energy of initiation and transformation), and the lover (the energy that connects one to others and the world), as well as the four immature patterns that interfere with masculine potential (divine child, oedipal child, trickster and hero).  King, Warrior, Magician, Lover is an exploratory journey that will help men and women reimagine and deepen their understanding of the masculine psyche.
Common People - An Anthology of...
Kit de Waal
Common People is the latest campaigning anthology from the publisher of bestsellers The Good Immigrant (70k copies sold to date) and Repeal the 8th.It features original work from some of our best-known writers: Kit De Waal, Malorie Blackman, Lisa McInerney, Cathy Rentzenbrink, Stuart Maconie, Louise Doughty, Damian Barr and Daljit Nagra among others.Kit de Waal's debut novel, My Name is Leon, was an international bestseller. It won Kerry Group Irish Novel of the Year 2017 and was shortlisted for the Costa First Book Award, the Desmond Elliott Prize and the British Book Awards 2017 Best Debut Book of the Year.Kit is one of the UK's most prominent voices calling for inclusivity and working-class representation in the arts, and has established a creative writing scholarship at Birkbeck University.Almost half of all authors, writers and
translators in the UK come from professional, middle-class backgrounds,
compared with just 10 per cent of those from working-class backgrounds. This
anthology addresses that pressing imbalance.There will be events across the UK on publication, including all the major literary festivals.For fans of The Good Immigrant, Know Your Place: Essays on the Working Class by the Working Class, Brit(ish) by Afua Hirsch.
Homer and Classical Philology
Friedrich Nietzsche was an extremely popular 19th century German philosopher.  Nietzsche wrote on a variety of topics including religion, morality, science, and modern culture.  Many scholars still study Nietzsche's works and ideology today.  This edition of Homer and Classical Philology includes a table of contents.

redpajama_set_name: RedPajamaCommonCrawl


text:
Slayer, Testament and Carcass Live at Boston's House of Blues
Slayer (all photos and words by Ben Stas)
Tom Araya is a man of few words, between songs at least. Slayer's grizzled frontman barked lyrics at a frantic clip, but kept his addresses to the crowd to a polite minimum over the course of the band's nearly two-hour set on March 6 at House of Blues in Boston, Massachusetts. The sold-out crowd, which spanned from giddy teenagers in Metallica shirts to scene veterans sporting well-worn battle jackets, needed no hyping up, and Slayer were committed to giving them exactly what they came for in a brutally efficient, no-nonsense performance.
Slayer are currently on the road with fellow California thrash mainstays Testament and Britain's legendary Carcass, supporting their first studio effort since the death of founding guitarist and songwriter Jeff Hanneman. It's tempting to think of Repentless and its ensuing tour, also less founding drummer Dave Lombardo and leaving just Araya and guitarist Kerry King as original members, as representing a lesser version of the band. In some ways of course, it is, but King and Araya's decision to soldier on doesn't ring hollow from the stage. There was a reverence for Slayer's legacy apparent in how vicious and vital both the new songs and the classics sounded.
Araya's vocals have lost none of their coarse effectiveness after all these years, and King's wildly technical fretwork flowed with confidence. With erstwhile drummer Paul Bostaph standing in for Lombardo and Exodus' Gary Holt doing an admirable job in Hanneman's place, the rounded out lineup also has pedigree – and more importantly, chemistry. Standing front and center as the first notes of Repentless' title track rang out and the physically overwhelming blast of low-end punched its way forth, there was no mistaking this for some invalid version of Slayer. The sheer sonic magnitude of the set appeared to overpower and temporarily cripple the venue's rather robust PA system – a first in the many shows I've seen there.
Slayer's setlists aren't known for their unpredictability, but the guarantee of hearing "Raining Blood" and "South of Heaven" regardless of the night is certainly not a bad one. The latter remains a song so immortally badass as to spontaneously inspire a satellite mosh pit between three guys at the merch table. Repentless isn't going to become anyone's go-to Slayer album, but the selections slotted in here held up amid the fan favorites. After a 22-song set that delivered many of the essentials concluded with "Angel of Death" – and the stage backdrop switching up to bear a Hanneman tribute – it was hard to imagine anyone leaving unsatisfied.
That can be credited in part to the band's impeccable choice in tour-mates, as well. Carcass' melodic death was the genre outlier of the night, but the crossover appeal was evident in the room's excited buzz before they took the stage. The Liverpool foursome mounted a comeback more successful than most with 2013's excellent Surgical Steel, their first record since 1996, and ripped through four of its tracks along with pairs of Necroticism and Heartwork favorites in a furious opening set. Like Slayer, the band's lineup now features a pair of original members in bassist/vocalist Jeff Walker and guitarist Bill Steer along with newer additions on second guitar and drums, but hardly sounds hindered by it. Walker was in fine gravel-voiced form, and the band made the most of the unfortunately brief time constraints of their set.
If Carcass were the night's outliers in terms of sound, it was Testament's air-guitaring, mosh-encouraging antics that incorporated a bit of a rock and roll partying vibe absent from the show's first and last sets. Vocalist Chuck Billy was clearly having a blast playing ringleader while the band traversed its 30-plus year history, touching on six of its ten albums since 1987. Testament may have lacked the precision heaviness of Carcass and the grim intensity of the night's headliners, but the touch of levity offered by their raucous set brought balance.
While the future of Slayer beyond the promotion of Repentless remains to be seen, a tour with curation and performances as strong as this one wouldn't be a bad note to go out on. It's an early contender for Q1 2016's best package deal.
Filed Under: bstas, carcass, slayer, testament
Categories: Features, live reviews


The <HTML> tag is used to start the HTML documents and to make the browsers to understand it is a HTML documents.
The container for all other HTML elements is <html> tag.
<html> tag is denoted as the root of the HTML page.
<html> tag supports Global attributes.
"Manifest" attribute is not supported.
html tag – defines the root of the HTML document.
We can see the output content of html tag in the web browser.


text:
Q: Text border - menu style
Can someone help me to get this kind of line, I have also seen it between categories.
I have tried like this but it is so close, I want it to be between categories.
#testborder{
border-right: 1px solid grey;
}
All answers welcome!

A: Next time you must give some code before asking. Or you may be not to have an answer
Html
<!DOCTYPE html>

    <title>MM Web Design</title>
    <link rel="stylesheet" href="css/all.css" />

bla bla

bla bla

bla bla

bla bla

bla bla

CSS
#nav {
    width: 100%;
    overflow: hidden;
    margin:0;
    padding:0;
    list-style:none;
}
#nav li {
    float: left;
    padding:  0 10px 0 9px;
    border-left: 1px solid #ccc;
}
#nav li:first-child {border: none;}


text:
title: "Learn how to use R for population genetics"
---

## Primers in population genetics for specific R packages

* [*adegenet*](http://adegenet.r-forge.r-project.org/)
* [*pegas*](http://ape-package.ird.fr/pegas.html)
* [*poppr*](http://grunwaldlab.github.io/Population_Genetics_in_R/)
* [*gstudio*](http://dyerlab.github.io/gstudio/)

## Other resources

* TBD

redpajama_set_name: RedPajamaGithub


text:
Aachen (franska Aix-la-Chapelle, nederländska Aken, på den lokala dialekten Öcher Platt Oche) är en stad i den tyska delstaten Nordrhein-Westfalen med cirka  invånare. Katedralen i Aachen utmärktes 1978 till första tyska kulturminnesmärke, och upptogs som andra kulturminnesmärke i Unescos världsarvslista.

Aachen är den västligaste tyska storstaden, och är känd för sin långa historia, sitt arkitektoniska och kulturella arv samt viktiga händelser som till exempel den årliga utdelningen av priset Karlspreis.

År 1890 översteg folkmängden i Aachen för första gången gränsen på 100 000, och staden blev därigenom en storstad enligt tyska standarder. Aachen ligger på den nordliga randen av Eifel. Större städer som ligger på ett avstånd av en timme med bil är Düsseldorf, Köln, Liège (Belgien), Mönchengladbach, Krefeld, Heerlen, Maastricht och Roermond (de tre senare i Nederländerna).

Aachen är hem till RWTH Aachen, en av de största och mest traditionsrika tekniska högskolorna i Europa, som sedan 2007 räknas till de tyska elituniversiteten.

För sin uppkomst har det moderna Aachen att tacka de varma svavelhaltiga hälsokällorna, vilka fortfarande är en av dess största inkomstkällor och årligen besöks av många badgäster. Den hetaste av källorna har en temperatur av 73,9 °C. Vattnet används mot reumatism, gikt, neuralgier och hudsjukdomar.

Historia

 Under första århundradet e.Kr. byggde romarna ett militärbad på platsen, där tidigare också kelter hade bott.
 Den frankiske kungen Pippin den lille uppförde på 700-talet en herrgård i Aachen.
 Hans son Karl den store ärvde kungariket och herrgården i Aachen och utvidgade den till ett palats, som han valde som sitt residens. Karl den store har också sin grav i domen, som tillhör världsarvet.
 1165 lät Fredrik Barbarossa saligförklara Karl den store och året därpå fick staden sina stadsrättigheter och blev också en fri Riksstad. Detta är skälet till att staden har en örn i sitt vapen.
 1656 förstördes praktiskt taget hela staden av en våldsam brand. I samband med återuppbyggnaden utvidgades kuranläggningen med diverse nöjesanordningar och staden blev därefter modebadet för Europas prominens.
 1668 slöts fred mellan Ludvig XIV och alliansen av England, Holland och Sverige i Aachen, där Frankrike tvingade återlämna alla sina erövringar i Spanien.
 1748 slöts även i Aachen fred efter det österrikiska arvföljdskriget.
 1794 blev staden i samband med den franska revolutionen fransk, men det varade endast till Wienkongressen 1815 då Aachen tilldelades konungariket Preussen.
 1818 möttes representanter för tidens stormakter i Aachen för att konferera. Under konferensen debatterades bland annat den interkontinentala slavhandeln.
 Omkring 1890 översteg invånarantalet 100 000, vilket gjorde Aachen till en storstad.
 En stor del av staden förstördes under andra världskriget och Aachen var den första större tyska staden som sedan intogs av amerikanska trupper.

Geografi
Aachen är den av de större städerna i Tyskland, som ligger längst västerut, belägen i regionen Euregio Maas-Rhein vid det så kallade Dreiländereck (en trelandspunkt mellan Tyskland, Nederländerna och Belgien) och vid den norra gränsen av området Eifel.

Den högsta punkten i stadsområdet ligger 410 meter över havet och befinner sig i sydöstra utkanterna av staden. Den lägsta punkten befinner sig 125 meter över havet, och ligger i norra delen av staden. Längden av stadsgränsen är 87,7 km, varav 23,8 km är gräns till Belgien, och 21,8 km gräns till Nederländerna. Den största bredden i nord-sydlig riktning är 21,6 km, och i öst-västlig riktning 17,2 km.

Befolkning
Staden Aachen hade 259 030 invånare i slutet av 2007.

Åldersstrukturen är som följer: 0–18 år: 16,1 %; 18–65 år: 67,2 %; från 65 år: 16,7 %. Snittåldern är 39,4 år (14 januari 2007).

Arbetslösheten var i maj 2007 11%, och andelen utlänningar 17,7%.

Storstadsområde
Många pendlar in till arbetsplatser och utbildningar i Aachenområdets centrala delar, och hela pendlingsområdet omfattar, förutom Aachen, 14 andra städer och kommuner. Större förorter är bland annat Alsdorf, Eschweiler, Herzogenrath och Stolberg. Bland Aachens distrikt återfinns bland annat Aachen-Mitte, Brand, Eilendorf, Haaren, Kornelimünster/Walheim, Laurensberg och Richterich.

Aachen bildar tillsammans med Heerlenområdet på den nederländska sidan ett internationellt storstadsområde med ungefär 900 000 invånare.

Kommunikationer
I Aachens utkanter korsar motorvägen A44 som går från Düsseldorf till Liège och Bryssel i Belgien och A4 från Köln till Maastricht, Eindhoven och Amsterdam i Nederländerna.

Aachen har utöver Aachen Hauptbahnhof (huvudstationen) stationerna Aachen West och Aachen Rothe Erde liksom de mindre tåghållplatserna Aachen Schanz och Eilendorf. Snabbtåget Thalys från Köln till Paris stannar i Aachen, Liège och Bryssel. Även ICE International trafikerar sträckan Bryssel–Liège–Aachen–Köln och vidare till Frankfurt. Snabbtågen Thalys och ICE går växelvis varje timme. Aachen ligger därmed på en viktig förbindelse i den transeuropeiska järnvägsnätet. Däremot är den pendlarinriktade regionaltrafiken, framförallt tågen till Köln, ofta försenade. På den andra utgående järnvägssträckan till Düsseldorf går sedan ungefär 10 år ingen fjärrtrafik längre. Båda sträckor är dock fortfarande viktiga förbindelser för godstransport.

Cirka 30 kilometer från staden ligger Maastricht Aachen Airport 1 mil norr om Maastricht (Nederländerna).

Ekonomi
De viktigaste industrierna i Aachen är tillverkning av järnvägsfordon (Talbot), bildäck (Uniroyal och Continental), lampor (Philips) och glas (Vegla). Staden är också känd för tillverkning av godsaker, bland annat kakorna Aachener Printen. Zentis är känt för marmeladtillverkning. Även turismen är av betydelse för staden, med 4 000 hotellbäddar och 680 000 övernattningar per år. Forskning och utveckling blir allt viktigare, Ericsson, Ford, Matsushita och Mitsubishi har etablerat sig i Aachen på grund av det centrala läget i Europa. Även den välkända chokladfabriken Lindt har tillverkning och fabriksförsäljning i Aachen.

Sport
Aachen arrangerade ryttar-VM 1978 och 2006. TSV Alemannia Aachen spelar i fotbollens 4. Fußball-Regionalliga.

Utbildning
I Aachen ligger förutom Rheinisch-Westfälische Technische Hochschule (RWTH), även tre andra högskolor, vilket också avspeglar sig i förekomsten av olika avknoppningsföretag från högskolorna.

Vänorter
Aachen har följande vänorter:
  Arlington County, USA
  Halifax, West Yorkshire, Storbritannien
  Kostroma, Ryssland
  Montebourg, Frankrike
  Naumburg an der Saale, Tyskland
  Ningbo, Kina
  Reims, Frankrike
  Sarıyer, Turkiet
  Toledo, Spanien

Referenser

Externa länkar
 www.aachen.de (tyska)
 www.rwth-aachen.de (tyska)

Kommuner och städer i Städteregion Aachen

Gränsorter i Europa
Kurorter i Tyskland
Badorter i Tyskland


text:
Jakob Krause (* 1531/1532 in Zwickau; † 9. Juli 1586 in Dresden), andere Schreibweisen Krauße, Krauss und Kreuss, war ein deutscher Buchbinder der Renaissance.

Leben

Krause besuchte die von Petrus Plateanus geleitete Lateinschule und begann eine Buchbinderlehre bei Christoph Behem. In seinen Wanderjahren 1549/50 arbeitete er unter anderem bei Lazarus Bernhart in Annaberg, ferner werden Aufenthalte in Leipzig und Wittenberg angenommen. Entscheidend für seine Entwicklung wurde ein mehrjähriger Aufenthalt in Frankreich, wo er den mauresken Stil, die Arabesken und den floralen Stil kennenlernte, die für seine späteren Einbände charakteristisch sind. Bei seiner Rückkehr brachte er viele Arbeitsinstrumente aus Frankreich mit.

Ab 1561 arbeitete er in Augsburg als Meister (Meisterprüfung im Frühjahr 1561 bei Jakob Holl) und heiratete. 1561 besuchte er seine Geburtsstadt Zwickau, danach weilte er in Wittenberg zum Stempelkauf. Krause erlangte das Augsburger Bürgerrecht, das er bis kurz vor seinem Tod beibehielt. In Augsburg fertigte er Dedikationsbände mit Handvergoldung an, z. B. 1561 einen Geschenkband an den Zwickauer Rat, zwei Empfehlungsbände an den Herzog von Bayern 1563 und eine Abschiedsgabe an den Augsburger Rat 1564. Es gelang ihm im Jahre 1566, am 25. August, als Hofbuchbinder in die Dienste des Kurfürsten August in Dresden zu treten, wo er eine Werkstatt in der Kanzlei erhielt. Es war ihm erlaubt, auch private Aufträge anzunehmen. Einer seiner Gesellen war ca. 1566–1568 der ebenfalls aus Zwickau stammende Gregor Schenk der Jüngere, später Hofbuchbinder des Würzburger Fürstbischofs Julius Echter. Krauses Gehilfe von 1574 bis 1578, der ihm auch im Amt des Hofbuchbinders folgen sollte, war Kaspar Meuser.

In Dresden fertigte Krause etwa 1000 Einbände an, von denen ein großer Teil heute in der SLUB Dresden aufbewahrt wird. Krause gilt neben seinem Nachfolger Meuser als bedeutendster deutscher Buchbinder der Renaissance. Er war Namensgeber für den 1912 gegründeten Jakob Krauße-Bund der deutschen Buchbinder.

Literatur
 Karl Berling: Der Kursächsische Hofbuchbinder Jakob Krause. Hoffmann, Dresden 1897 (Digitalisat).

 Ilse Schunke: Krause-Studien. In: Zentralblatt für Bibliothekswesen. Beiheft 65, Harrassowitz, Leipzig 1932.
 Ilse Schunke: Leben und Werk Jakob Krauses. Insel-Verlag, Leipzig 1943.

 Angelika Pabel: Der runde Renaissance-Einband für Fürstbischof Julius Echter. In: Einbandforschung. 20, 2007, S. 27–36.

Weblinks

 Einband von Jakob Krause (PDF; 1,5 MB)
 Das Rundbuch des Fürstbischofs Julius Echter.
 Jakob Krause in der Einbanddatenbank

Einzelnachweise

Person (Dresden)
Buchbinder
Deutscher
Geboren im 16. Jahrhundert
Gestorben 1586
Mann


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