text:
\begin{document}

\title{Relieving and Readjusting Pythagoras}

\author{Victor Luo}\email{\textcolor{blue}{\href{mailto:victor.d.luo@williams.edu}{victor.d.luo@williams.edu}}}
\address{Department of Mathematics and Statistics, Williams College, Williamstown, MA 01267}

\author{Steven J. Miller}\email{\textcolor{blue}{\href{mailto:sjm1@williams.edu}{sjm1@williams.edu}},  \textcolor{blue}{\href{Steven.Miller.MC.96@aya.yale.edu}{Steven.Miller.MC.96@aya.yale.edu}}}
\address{Department of Mathematics and Statistics, Williams College, Williamstown, MA 01267}

\subjclass[2000]{46N30 (primary), 62F03, 62P99  (secondary).}

\keywords{Pythagorean Won-Loss Formula, Weibull Distribution, Hypothesis Testing}

\date{\today}

\thanks{The second named author was partially supported by NSF grant DMS0970067. We thank Kevin Dayaratna, Bernhard Klingenberg and Jeffrey Miller for helpful comments and code over the years.}

\begin{abstract}
Bill James invented the Pythagorean expectation in the late 70's to predict a baseball team's winning percentage knowing just their runs scored and allowed. His original formula estimates a winning percentage of $\rs^2/(\rs^2+\ra^2)$, where $\rs$ stands for runs scored and $\ra$ for runs allowed; later versions found better agreement with data by replacing the exponent 2 with numbers near 1.83. Miller and his colleagues provided a theoretical justification by modeling runs scored and allowed by independent Weibull distributions. They showed that a single Weibull distribution did a very good job of describing runs scored and allowed, and led to a predicted won-loss percentage of $(\rso-1/2)^\gamma / ((\rso-1/2)^\gamma + (\rao-1/2)^\gamma)$, where $\rso$ and $\rao$ are the observed runs scored and allowed and $\gamma$ is the shape parameter of the Weibull (typically close to 1.8). We show a linear combination of Weibulls more accurately determines a team's run production and increases the prediction accuracy of a team's winning percentage by an average of about 25\% (thus while the currently used variants of the original predictor are accurate to about four games a season, the new combination is accurate to about three). The new formula is more involved computationally; however, it can be easily computed on a laptop in a matter of minutes from publicly available season data. It performs as well (or slightly better) than the related Pythagorean formulas in use, and has the additional advantage of having a theoretical justification for its parameter values (and not just an optimization of parameters to minimize prediction error).
\end{abstract}

\maketitle

\tableofcontents

\section{Introduction}

The Pythagorean Win/Loss Formula, also known as the Pythagorean formula or Pythagorean expectation, was invented by Bill James in the late 1970s to use a team's observed runs scored and allowed to predict their winning percentage. Originally given by \be {\rm Won-Loss\ Percentage} \ = \ \frac{\rs^2}{\rs^2+\ra^2}, \ee with $\rs$ the runs scored and $\ra$ the runs allowed, it earned its name from the similarity of the denominator to the sums of squares in the Pythagorean formula from geometry.\footnote{Though of course the more natural shape in baseball is the diamond, save for some interesting stadium features, such as the triangle in Fenway Park.} Later versions found better agreement by replacing the exponent 2 with numbers near 1.83, leading to an average error of about three to four games per season.

The formula is remarkably simple, requiring only the runs scored and allowed by a team in a season, and the calculation (even with the improved exponent) is easily done on any calculator or phone. It is one of the most commonly listed expanded statistics on websites. One reason for its prominence is its accuracy in predicting a team's future performance through a simple calculation and not through computationally intense simulations. Additionally, it allows sabermetricians and fans to assess a manager's impact on a team, and estimate the value of new signings by seeing how their presence would change the predictions.

Because of its widespread use and utility, it is very desirable to have improvements. In his senior thesis, the first named author, supervised by the second named author, explored various attempted improvements to the Pythagorean formula. These included replacing the observed runs scored and allowed each game with adjusted numbers, with the adjustments coming from a variety of sources (such as ballpark effects, game state\footnote{For example, if a team is up by a large amount late in a game, they frequently use weaker relief pitchers and rest some starters, while the trailing team makes similar moves; thus the offensive productions from this point onward may not be indicative of the team's true abilities and a case can be made to ignore such data.}, WHIP, ERA+, and WAR of the pitcher, ...).

As these led to only minor improvements\footnote{The only adjusted formula that was at least on par or very near the accuracy of the original Pythagorean W/L Formula was that of ballpark factor.} (see \cite{Luo} for a detailed analysis of these and other adjustments), we turned our attention to the successful theoretical model used by Miller and his colleagues \cite{DaMil1,DaMil2,Mil,MCGLP}, where it was assumed runs scored and allowed were independently drawn from Weibull distributions with the same shape parameter. Recall the three parameter Weibull density is given by \be \twocase{f(x;\alpha,\beta,\gamma) \ = \ }{\frac{\gamma}{\alpha}\ ((x-\beta)/\alpha)^{\gamma-1}\ e^{- ((x-\beta)/\alpha)^{\gamma}}}{if $x \ge \beta$}{0}{otherwise.} \ee The effect of $\alpha$ is to control the spread of the output, while $\beta$ translates the distribution. The most important parameter is $\gamma$, which controls the shape. See Figure \ref{fig:weibullplots} for some plots.

\begin{center}
\begin{figure}
\includegraphics[scale=0.65]{weibulldist-eps-converted-to}
\caption{\label{fig:weibullplots} The varying distributions of the Weibull family with $\alpha=1$ and $\beta=0$.}

\end{figure}
\end{center}

Their success is due to the fact that the three parameter Weibull is a very flexible family of distributions, capable of fitting many one hump distributions, including to a statistically significant degree the observed runs scored and allowed data. Miller chose to use Weibulls for two reasons. First, they lead to double integrals for the probabilities that can be evaluated in closed form. This is extremely important if we desire a simple expression such as the one posited by James (see \cite{HJM} for alternative simple formulas). Second, in addition to being flexible, special values of the Weibulls correspond to well-known distributions ($\gamma = 1$ is an exponential, while $\gamma = 2$ is the Rayleigh distribution).

The goal of this paper is to show that one can significantly improve the predictive power if instead of modeling runs scored and allowed as being drawn from independent Weibulls, we instead model them as being drawn from linear combinations of independent Weibulls. The advantage of this approach is that we are still able to obtain tractable double integrals which can be done in closed form. There is a cost, however, as now more analysis is needed to find the parameters and the correct linear combinations. While this results in a more complicated formula than the standard variant of James' formula, it is well worth the cost as on average it is better by one game per season (thus a typical error is 3 games per team per year, as opposed to 4 which is the typical result in the current formula). Comparing it to \url{baseball-reference.com}'s calculated Expected WL from 1979 to 2013, we find that the linear combination of Weibulls is approximately .06 of a game better, a difference is not statistically significant. However, there are noticeable trends that appear in certain eras.

\section{Theoretical Calculations} \label{sec:theoretical}

\subsection{Preliminaries}

It is important to note that we assume that runs scored and allowed are taken from continuous, not discrete, distributions. This allows us to deal with continuous integrals rather than discrete sums, which most of the time leads to easier calculations. While a discrete distribution would probably more effectively map runs in baseball, the assumption of drawing runs from a continuous distribution allows for more manageable calculations, and is a very sensible estimate of those runs observed. It also will lead to closed form expressions, which are much easier to work with and allow us to avoid having to resort to simulations.

The Weibulls lead to significantly easier calculations because if we have a random variable $X$ chosen from a Weibull distribution with parameters $\alpha$, $\beta$, and $\gamma$, then $X^{1/\gamma}$ is exponentially distributed with parameter $\alpha^\gamma$; thus, a change of variables yields a simpler integral of exponentials, which can be done in closed form (see Appendix 9.1 in \cite{MCGLP} for details).

In all arguments below we always take $\beta=-1/2$, though we often write $\beta$ to keep the discussion more general for applications to other sports. The reason we do this is that we use procedures such as the Method of Least Squares to find the best fit parameters, and this requires binning the observed runs scored and allowed data. As baseball scores are discrete, there are issues if these values occur at the boundary of bins; it is much better if they are at the center. By taking $\beta=-1/2$ we break the data into bins
\begin{equation} \label{bins}
\left[-\frac{1}{2},\ \frac{1}{2}\right),\ \ \ \left[\frac{1}{2},\ \frac{3}{2}\right),\ \ \ \left[\frac{3}{2},\ \frac{5}{2}\right),\ \ \ \cdots.
\end{equation}

Our final assumption is that runs scored and runs allowed are independent. This obviously cannot be true, as a baseball game never ends in a tie. For example, if the Orioles and the Red Sox are playing and the O's score 5 runs, then the Sox cannot score 5. Statistical analyses support this hypothesis; see the independence tests with structural zeros in \cite{Mil} or Appendix 9.2 in \cite{MCGLP} for details.

We end this subsection with the mean and the variance of the Weibull. The calculation follows from standard integration (see the expanded version of \cite{Mil} or Appendix 9.1 in \cite{MCGLP} for a proof of the formula for the mean; the derivation of the variance follows in a similar fashion).

\begin{lemma} \label{weibullmv} Consider a Weibull with parameters $\alpha, \beta, \gamma$. The mean, $\mu_{\alpha,\beta,\gamma}$, equals
\be
\mu_{\alpha,\beta,\gamma}\ = \ \alpha \Gamma (1+\gamma^{-1})+\beta \label{weibullmean}
\ee
while the variance, $\sigma^2_{\alpha,\beta,\gamma}$, is
\be
\sigma^2_{\alpha,\beta,\gamma}\ = \ \alpha^2 \Gamma(1+2\gamma^{-1})-\alpha^2\Gamma(1+\gamma^{-1})^2,
\ee
where $\Gamma(x)$ is the Gamma function, defined by $\Gamma(x)=\int_0^\infty e^{-u}u^{x-1}du$.
\end{lemma}

\subsection{Linear Combination of Weibulls}

We now state and prove our main result for a linear combination of two Weibulls, and leave the straightforward generalization to combinations of more Weibulls to the reader. The reason such an expansion is advantageous and natural is that, following \cite{Mil}, we can integrate pairs of Weibulls in the regions needed and obtain simple closed form expressions. The theorem below also holds if $\gamma < 0$; however, in that situation the more your runs scored exceeds your runs allowed, the worse your predicted record due to the different shape of the Weibull (in all applications of Weibulls in survival analysis, the shape parameter $\gamma$ must be positive).

\begin{theorem} \label{combweib}
Let the runs scored and allowed per game be two independent random variables drawn from linear combinations of independent Weibull distributions with the same $\beta$'s and $\gamma$'s. Specifically, if $W(t;\alpha,\beta,\gamma)$ represents a Weibull distribution with parameters $(\alpha,\beta,\gamma)$,  and we choose non-negative weights\footnote{If we had more terms in the linear combination, we would simply choose non-negative weights summing to 1.} $0\le c_i, c_j' \le 1$  (so $c_1 + c_2 = 1$ and $c_1'+c_2' = 1$), then the density of runs scored, $X$ is \be f(x;\alpha_{{\rm RS}_1},\alpha_{{\rm RS}_2}, \beta, \gamma,c_1, c_2) \ = \ c_1 W(x;\alpha_{{\rm RS}_1},\beta,\gamma)+c_2 W(\alpha_{{\rm RS}_2},\beta,\gamma)\ee and runs allowed, $Y$, is \be f(y;\alpha_{{\rm RA}_1},  \alpha_{{\rm RA}_2}, \beta, \gamma,c_1', c_2') \ = \ c_1' W(y;\alpha_{{\rm RA}_1},\beta,\gamma)+c_2'W(\alpha_{{\rm RA}_2},\beta,\gamma).\ee In addition, we choose $\alpha_{{\rm RS}_1}$ and $\alpha_{{\rm RS}_2}$ so that the mean of $X$ is $\rso$ and choose $\alpha_{{\rm RA}_1}$ and $\alpha_{{\rm RA}_2}$ such that the mean of $Y$ is $\rao$. For $\gamma>0$, we have
\bea & & {\rm Won-Loss\ Percentage}(\alpha_{{\rm RS}_1}, \alpha_{{\rm RS}_2}, \alpha_{{\rm RA}_1}, \alpha_{{\rm RA}_2},\beta,\gamma,c_1, c_2, c_1',c_2') \nonumber\\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  =\  c_1 c_1' \frac{\alpha_{{\rm RS}_1}^\gamma}{\alpha_{{\rm RS}_1}^\gamma+\alpha_{{\rm RA}_1}^\gamma} +c_1 c_2'\frac{\alpha_{{\rm RS}_1}^\gamma}{\alpha_{{\rm RS}_1}^\gamma+\alpha_{{\rm RA}_2}^\gamma} \nonumber\\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ +c_2 c_1' \frac{\alpha_{{\rm RS}_2}^\gamma}{\alpha_{{\rm RS}_2}^\gamma +\alpha_{{\rm RA}_1}^\gamma} +c_2 c_2'\frac{\alpha_{{\rm RS}_2}^\gamma}{\alpha_{{\rm RS}_2}^\gamma+\alpha_{{\rm RA}_2}^\gamma} \nonumber\\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = \ \sum_{i=1}^2 \sum_{j=1}^2 c_i c_j' \frac{\alpha_{{\rm RS}_i}^\gamma}{\alpha_{{\rm RS}_i}^\gamma+\alpha_{{\rm RA}_j}^\gamma}. \eea
\end{theorem}

\begin{proof}
As the means of $X$ (runs scored) and $Y$ (runs allowed) are $\rso$ and $\rao$, respectively, and the random variables are drawn from linear combinations of independent Weibulls, by Lemma \ref{weibullmv}
\begin{align}
\rso &\ =\ c_1(\alpha_{{\rm RS}_1}\Gamma(1+\gamma^{-1})+\beta)+(1-c_1)(\alpha_{{\rm RS}_2}\Gamma(1+\gamma^{-1})+\beta) \nonumber \\
\rao &\ =\ c_1'(\alpha_{{\rm RA}_1}\Gamma(1+\gamma^{-1})+\beta)+(1-c_1')(\alpha_{{\rm RA}_2}\Gamma(1+\gamma^{-1})+\beta).
\end{align}

We now calculate the probability that $X$ exceeds $Y$. We constantly use the fact that the integral of a probability density is 1. We need the two $\beta$ and the two $\gamma$'s to be equal in order to obtain closed form expressions.\footnote{If the $\beta$'s are differen then in the integration below we might have issues with the bounds of integration, while if the $\gamma$'s are unequal we get incomplete Gamma functions, though for certain rational ratios of the $\gamma$'s these can be done in closed form.} We find
\begin{align}
& {\rm Prob}(X>Y)  = \int_{x=\beta}^\infty\int_{y=\beta}^x f(x; \alpha_{{\rm RS}_1},\alpha_{{\rm RS}_2},\beta,\gamma, c_1, c_2)f(y; \alpha_{{\rm RA}_1},\alpha_{{\rm RA}_2},\beta,\gamma, c_1',c_2')dydx  \nonumber \\
 &=\  \sum_{i=1}^2\sum_{j=1}^2 \int_{x=0}^\infty \int_{y=0}^x c_ic_j' \frac{\gamma}{\alpha_{{\rm RS}_i}}\left(\frac{x}{\alpha_{{\rm RS}_i}}\right)^{\gamma-1}e^{-(\frac{x}{\alpha_{{\rm RS}_i}})^\gamma}\frac{\gamma}{\alpha_{{\rm RA}_j}}\left(\frac{x}{\alpha_{{\rm RA}_j}}\right)^{\gamma-1}e^{-(\frac{x}{\alpha_{{\rm RA}_j}})^\gamma}dydx \nonumber \\
& = \sum_{i=1}^2\sum_{j=1}^2 c_ic_j' \int_{x=0}^\infty \frac{\gamma}{\alpha_{{\rm RS}_i}} \left( \frac{x}{\alpha_{{\rm RS}_i}} \right)^{\gamma-1}e^{-(\frac{x}{\alpha_{{\rm RS}_i}})^\gamma}\left[ \int_{y=0}^x \frac{\gamma}{\alpha_{{\rm RA}_j}} \left( \frac{y}{\alpha_{{\rm RA}_j}} \right)^{\gamma-1} e^{-(\frac{y}{\alpha_{{\rm RA}_j}})^\gamma} dy \right] dx \nonumber \\%
 &=\  \sum_{i=1}^2\sum_{j=1}^2 c_ic_j'  \int_{x=0}^\infty \frac{\gamma}{\alpha_{{\rm RS}_i}} \left( \frac{x}{\alpha_{{\rm RS}_i}} \right)^{\gamma-1}e^{-(\frac{x}{\alpha_{{\rm RS}_i}})^\gamma} \ast \left[ 1- e^{-(\frac{x}{\alpha_{{\rm RA}_j}})^\gamma} \right] dx \nonumber \\
 &=\  \sum_{i=1}^2\sum_{j=1}^2 c_ic_j'  \left[ 1- \int_{x=0}^\infty \frac{\gamma}{\alpha_{{\rm RS}_i}} \left( \frac{x}{\alpha_{{\rm RS}_i}}\right)^{\gamma-1} e^{-(\frac{x}{\alpha_{{\rm RS}_i}})^\gamma-(\frac{x}{\alpha_{{\rm RA}_j}})^\gamma} dx\right]. \label{eqqq}
\end{align}

We set
\begin{align}
\frac{1}{\alpha_{ij}^\gamma}\ = \ \frac{1}{\alpha_{{\rm RS}_i}^\gamma}+\frac{1}{\alpha_{{\rm RA}_j}^\gamma}\ = \ \frac{\alpha_{{\rm RS}_i}^\gamma+\alpha_{{\rm RA}_j}^\gamma}{\alpha_{{\rm RS}_i}^\gamma \alpha_{{\rm RA}_j}^\gamma} \nonumber \\ \nonumber
\end{align}
for $1\leq i,j\leq 2$, so that \eqref{eqqq} becomes
\begin{align}
& \sum_{i=1}^2\sum_{j=1}^2 c_ic_j'  \left[ 1- \int_{x=0}^\infty \frac{\gamma}{\alpha_{{\rm RS}_i}}\left( \frac{x}{\alpha_{{\rm RS}_i}}\right)^{\gamma-1}e^{-(\frac{x}{\alpha_{ij}})^\gamma} dx \right]  \nonumber \\
&=\ \sum_{i=1}^2\sum_{j=1}^2 c_ic_j' \left[ 1- \frac{\alpha_{ij}^\gamma}{\alpha_{{\rm RS}_i}^\gamma} \int_{x=0}^\infty \frac{\gamma}{\alpha_{ij}} \left( \frac{x}{\alpha_{ij}} \right)^{\gamma-1}e^{-(\frac{x}{\alpha_{ij}})^\gamma} dx \right] \nonumber \\

&=\  \sum_{i=1}^2 \sum_{j=1}^2 c_ic_j' \left[1-\frac{\alpha_{ij}^\gamma}{\alpha_{{\rm RS}_i}^\gamma} \right] \nonumber \\
&=\  \sum_{i=1}^2\sum_{j=1}^2 c_i c_j'\left[1-\frac{1}{\alpha_{{\rm RS}_i}^\gamma}\ast \frac{\alpha_{{\rm RS}_i}^\gamma\alpha_{{\rm RA}_j}^\gamma}{\alpha_{{\rm RS}_i}^\gamma+\alpha_{{\rm RA}_j}^\gamma}\right] \nonumber \\
&=\  \sum_{i=1}^2\sum_{j=1}^2 c_i c_j'\left[\frac{\alpha_{{\rm RS}_i}^\gamma}{\alpha_{{\rm RS}_i}^\gamma+\alpha_{{\rm RA}_j}^\gamma}\right] \nonumber \\
&=\  c_1c_1' \frac{\alpha_{{\rm RS}_1}^\gamma}{\alpha_{{\rm RS}_1}^\gamma+\alpha_{{\rm RA}_1}^\gamma} +c_1c_2' \frac{\alpha_{{\rm RS}_1}^\gamma}{\alpha_{{\rm RS}_1}^\gamma+\alpha_{{\rm RA}_2}^\gamma} \nonumber \\
&\indent +c_2c_1' \frac{\alpha_{{\rm RS}_2}^\gamma}{\alpha_{{\rm RS}_2}^\gamma +\alpha_{{\rm RA}_1}^\gamma}
+c_2c_2' \frac{\alpha_{{\rm RS}_2}^\gamma}{\alpha_{{\rm RS}_2}^\gamma+\alpha_{{\rm RA}_2}^\gamma},
\end{align} completing the proof of Theorem \ref{combweib}.\end{proof}

\section{Curve Fitting}

\subsection{Theory}

We now turn to finding the values of the parameters leading to the best fit. We require $\beta = -1/2$ (for binning purposes), but otherwise the parameters ($\alpha_{{\rm RS}_1}, \alpha_{{\rm RS}_2}$, $\alpha_{{\rm RA}_1}$, $\alpha_{{\rm RA}_2}$, $\gamma$, $c_1$, $c_2$, $c_1'$, $c_2'$) are free.\footnote{Subject to, of course, $0 \le c_i, c_j' \le 1$ and $c_1+c_1=c_1'+c_2'=1$.} Our first approach was to use the Method of Moments, where we compute the number of moments equal to the number of parameters. Unfortunately the resulting equations were too involved to permit simple solutions for them in terms of the observed data; for completeness they are given in Appendix \ref{sec:AppMoment} (or see \cite{Luo}). We thus turned to the Method of Least Squares (though one could also do an analysis through the Method of Maximum Likelihood).

We looked at the 30 teams of the entire league from the 2004 to 2012 season. We display results from the 2011, but the results from any other season are similar readily available (see \cite{Luo}). We implemented the Method of Least Squares using the bins in \eqref{bins},  which involved minimizing the sum of squares of the error of the runs scored data plus the sum of squares of the error of the runs allowed data. There were seven free parameters: $\alpha_{{\rm RS}_1}$, $\alpha_{{\rm RS}_2}$, $\alpha_{{\rm RA}_1}$, $\alpha_{{\rm RA}_2}$, $\gamma$, $c_1$, and $c_1'$. Letting Bin$(k)$ be the $k$\textsuperscript{th} bin of $\eqref{bins}$, $\rso(k)$ and $\rao(k)$ represent the observed number of games with number of runs scored and allowed in Bin$(k)$, and $A(\alpha_1,\alpha_2,\beta,\gamma,c_1,k)$ denote the area under the linear combination of two Weibulls with parameters $(\alpha_1,\alpha_2,\beta,\gamma,c_1)$ in Bin$(k)$, then for each team we found the values of $(\alpha_{{\rm RS}_1},\alpha_{{\rm RS}_2},\alpha_{{\rm RA}_1},\alpha_{{\rm RA}_2}, \gamma, c_1, c_1')$ that minimized
\begin{align}
& \sum_{k=1}^{\textnormal{Num. Bins}} (\rso(k)-\#\textnormal{Games}\ast A(\alpha_{{\rm RS}_1},\alpha_{{\rm RS}_2},-.5,\gamma,c_1,k))^2 \nonumber \\ &\indent +\sum_{k=1}^{\textnormal{Num. Bins}} (\rao(k)-\#\textnormal{Games}\ast A(\alpha_{{\rm RA}_1},\alpha_{{\rm RA}_2},-.5,\gamma,c_1',k))^2 .
\end{align}

\subsection{Results}

For each team, we found the best fit linear combination of Weibulls. In Figure \ref{leastsquarestable}, we compared the predicted wins, losses, and won-loss percentage with the observed ones.

\begin{figure}[h!]
\includegraphics[scale=0.5]{leastsquarestable-eps-converted-to}
\caption{Results for the 2011 season using Method of Least Squares. \label{leastsquarestable}}
\end{figure}

The code used is available in \cite{Luo}. Using the Method of Least Squares, the mean $\gamma$ over all 30 teams is 1.83 with a standard deviation of 0.18 (the median is 1.79). We can see that the exponent 1.83, considered as the best exponent, is clearly within the region of one standard deviation from the mean $\gamma$. Considering the absolute value of the difference between observed and predicted wins,  we have a mean of 2.89 with a standard deviation of 2.34 (median is 2.68). Without considering the absolute value, the mean is 0.104 with a standard deviation of 3.75 (and a median of 0.39). We only concern ourselves with the absolute value of the difference, as this really tells how accurate our predicted values are.

These values are significant improvements on those obtained when using a single Weibull distribution to predict runs (which essentially reproduces James' original formula, though  with a slightly different exponent), which produces a mean number of games off of 4.43 with standard deviation 3.23 (and median 3.54) in the absolute value case. We display the results over seasons from 2004 to 2012 in Figure \ref{sidebyside}. It is apparent that the linear combination of Weibulls better estimates teams' win/loss percentage; in fact, it is over one game better at estimating than the single Weibull! The mean number of games off for a single Weibull from 2004 to 2012 was 4.22 (with a standard deviation of 3.03), while that of the linear combination of Weibulls was 3.11 (with a standard deviation of 2.33). In addition, there is less standard deviation in the estimates. Thus, it appears that the linear combination of Weibulls provides a much tighter, better estimate than the single Weibull does.

\begin{figure}[h!]
\includegraphics[scale=0.4]{sidebyside-eps-converted-to}
\caption{Mean number of games off (with standard deviation) for single Weibull and linear combination of Weibulls from 2004-2012. \label{sidebyside}}
\end{figure}

To further demonstrate how accurate the quality of the fit is, we compare the best fit linear combination of Weibulls of runs scored and allowed with those observed of the 2011 Seattle Mariners in Figure \ref{mariners}; we can see that the fit is visually very good. Of course, the fit \emph{cannot} be worse, as we can always set $c_1 = 0=c_1'$; however, we can see the linear combination of Weibulls does a better job tracking the shape of the runs scored.

\begin{figure}
\centering
\begin{subfigure}{.53\textwidth}
  \centering
  \includegraphics[width=.8\linewidth]{singleseattle-eps-converted-to}
  \caption{Single Weibull mapping runs scored and \\ allowed.}
  \label{fig:sub1}
\end{subfigure}
\begin{subfigure}{.53\textwidth}
  \centering
  \includegraphics[width=.8\linewidth]{doubleseattle-eps-converted-to}
  \caption{Linear Combination of Weibulls mapping runs \\ scored and allowed.}
  \label{fig:sub2}
\end{subfigure}
\caption{Comparison of best fit linear combination of Weibulls versus single Weibull for runs scored (top) and allowed (bottom) for the 2011 Seattle Mariners against the observed distribution of scores.}
\label{mariners}
\end{figure}

We then performed an independent two-sample t-test with unequal variances in $R$ using the t.test command to see if the difference between the games off determined by the single Weibull and those by linear combinations of Weibulls is statistically significant in Figure \ref{test}. With a $p$-value less than 0.01 and a 95\% confidence interval that does not contain 0, we can see that the difference is in fact statistically significant.

\begin{figure}[h!]
\includegraphics[scale=0.7]{ttest-eps-converted-to}
\caption{t-test to determine whether the difference between the games off determined by the single Weibull and those by linear combinations of Weibulls is statistically significant.  \label{test}}
\end{figure}

In addition, we compared the mean number of games off of \url{baseball-reference.com}'s Pythagorean Win-Loss statistic (pythWL) and those of the linear combination of Weibulls from 1979 to 2013. Originally, we used ESPN's ExWL statistic\footnote{See the bottom of the page \url{http://espn.go.com/mlb/stats/rpi/_/year/2011}.} which used an exponent of 2; however, ESPN only went down to the year 2002, and it has been shown that using the exponent 1.83 is more accurate than using ESPN's exponent of 2. Using \url{baseball-reference.com}'s Pythagorean Win-Loss statistic (pythWL) to obtain more data (\url{baseball-reference.com} allowed us to go all the way down to 1979, rather than just 2002, which ESPN gives), the pythWL statistic\footnote{At \url{http://www.sports-reference.com/blog/baseball-reference-faqs/} see the section ``What is Pythagorean Winning Percentage?".} is calculated as
\begin{center}
$({\text{runs scored}^{1.83}})/({\text{runs scored}^{1.83}+\text{runs allowed}^{1.83}})$.
\end{center}

We display the results of our comparisons in Figure \ref{espnplot}. The mean number of games off for the pythWL statistic was 3.09 with a standard deviation of 2.26, numbers only slightly worse than those of the linear combination of Weibulls (mean of 3.03 with standard deviation of 2.21). So, we can see that the linear combination of Weibulls is doing, on average, about .06 of a game better than the pythWL statistic. We performed an independent two-sample t-test with unequal variances in $R$ using the t.test command to see if the difference between the games off determined by the pythWL statistic and the linear combinations of Weibulls is statistically significant in Figure \ref{ttest2}. With a very large $p$-value, we fail to reject the null hypothesis, suggesting that the difference in mean number of games off is not in fact statistically significant. We also display a plot (Figure \ref{difference}) that models the difference in the number of games between the pythWL statistic and the linear combination of Weibulls; it seems to be a constant positive value for the most part, suggesting that the linear combination of Weibulls is doing slightly better than the pythWL statistic.

\begin{figure}[h!]
\includegraphics[scale=0.46]{espnplot-eps-converted-to}
\caption{Mean number of games off (with standard deviation) for \url{baseball-reference.com}'s pythWL statistic and linear combination of Weibulls from 1979-2013. \label{espnplot}}
\end{figure}

Looking at Figure \ref{difference} more closely, we can see that there are parts/eras of the graph in which the pythWL statistic does better, and parts where the linear combination of Weibulls does better. In the era from 1979-1989, the pythWL statistic is more accurate, beating the linear combination of Weibulls in 7 out of the 11 years. However, from 1990 to 2013, the linear combination of Weibulls wins in 15 out of the 24 years, and does so by around 0.3 games in those years. Furthermore, when the pythWL statistic does beat the linear combination of Weibulls in the years from 1990 to 2013, it does so by around 0.25 games, including the point at 2004, which seems very out of the ordinary; without this point, the pythWL statistic wins by about .2 games in the years between 1990 and 2013 that it does beat the linear combination of Weibulls. Thus, in more recent years, it may make more sense to use the linear combination of Weibulls. In addition, with respect to the standard deviation of number of games off of the pythWL statistic (2.26) and the linear combination of Weibulls (2.21), we can see that the linear combination of Weibulls provides on average a tighter fit, i.e., there is less fluctuation in the mean number of games off for each team in each year (from 1990 to 2013, the pythWL statistic standard deviation in games off is 2.34 while that of the linear combination of Weibulls is 2.22, so we again see that the linear combination does noticeably better in recent years).

It is important to note that the pythWL statistic just takes the functional form of the Pythagorean Win/Loss Formula with an exponent ($\gamma$) of 1.83, while we give theoretical justification for our formula.

\begin{figure}[h!]
\includegraphics[scale=0.6]{ttest2-eps-converted-to}
\caption{t-test to determine whether the difference between the games off determined by ESPN ExWL and those by linear combinations of Weibulls is statistically significant.  \label{ttest2}}
\end{figure}

\begin{figure}[h!]
\includegraphics[scale=0.4]{difference-eps-converted-to}
\caption{Difference in mean number of games off for \url{baseball-reference.com}'s pythWL statistic and linear combination of Weibulls from 1979-2013. \label{difference}}
\end{figure}

We also performed $\chi^2$ tests to determine the goodness of fit to see how well the linear combination of Weibulls maps the observed data, and whether runs scored and allowed are independent. We used the bins as in \eqref{bins} and test statistic
\begin{align}
&\sum_{k=1}^{\textnormal{\# Bins}} \frac{(\rso(k)-\# \textnormal{Games} \ast  A(\alpha_{{\rm RS}_1},\alpha_{{\rm RS}_2}, -.5,\gamma,c_1,k))^2}{\# \textnormal{Games} \ast  A(\alpha_{{\rm RS}_1},\alpha_{{\rm RS}_2},-.5,\gamma,c_1,k)} \nonumber \\
&\indent + \sum_{k=1}^{\textnormal{\# Bins}} \frac{(\rao(k)-\# \textnormal{Games} \ast  A(\alpha_{{\rm RA}_1},\alpha_{{\rm RA}_2}, -.5,\gamma,c_1',k))^2}{\# \textnormal{Games} \ast  A(\alpha_{{\rm RA}_1},\alpha_{RS_A},-.5,\gamma,c_1',k)}
\end{align}
for the goodness of fit tests, with $2\ast (\# \textnormal{Bins} -1)-1-7 = 16$ degrees of freedom, the factor of 7 coming from estimating 7 parameters, namely $\alpha_{{\rm RS}_1}$, $\alpha_{{\rm RS}_2}$, $\alpha_{{\rm RA}_1}$, $\alpha_{{\rm RA}_2}$, $\gamma$, $c_1$, and $c_1'$. We did not estimate $\beta$, as we took it to be -.5. Having 16 degrees of freedom gives critical threshold values of 26.3 (at the 95\% level) and 32.0 (at the 99\% level). However, since there are multiple comparisons being done (namely 30 for the different teams), we use a Bonferroni adjustment and obtain critical thresholds of 37.7 (95\%) and 42.5 (99\%). From the first column of Figure \ref{independence}, all the teams fall within the unadjusted 99\% threshold, with the exception of the Texas Rangers (just barely!), who easily fall into the Bonferroni adjusted 95\% threshold. Therefore, the observed data closely follows a linear combination of Weibulls with the proper estimated parameters.

Since the test for independence of runs scored and allowed requires that the row and column of the contingency table have at least one non-zero entry, the bins used to bin the runs score and allowed were
\be
[0,1)\ \cup\ [1,2) \ \cup\ \cdots\ \cup\ [9,10)\ \cup\ [11,\infty).
\ee

We use integer endpoints because we are using the observed runs from games. We have a 12 by 12 contingency table with zeroes along the diagonal, since runs scored and allowed can never be equal. This leads to an incomplete 12 by 12 contingency table with $(12-1)^2-12=109$ degrees of freedom; constructing a test requires the use of structural zeroes. The theory behind tests using structural zeroes can be seen in \cite{Mil} or Appendix 9.2 of \cite{MCGLP}. We observe that 109 degrees of freedom give critical threshold values of 134.37 (at the 95\% level) and 146.26 (at the 99\% level). Again, since we are doing multiple comparisons, we use a Bonferroni adjustment, obtaining critical thresholds of 157.68 (95\%) and 166.45 (99\%). From the second column of Figure \ref {independence}, all the teams fall within the 99\% threshold, with the exception of the Los Angeles Angels (just barely!), who easily fall into the Bonferroni adjusted 95\% threshold. Thus, runs scored and allowed are acting as though they are statistically independent.

\begin{figure}[h!]
\includegraphics[scale=0.51]{independence-eps-converted-to}
\caption{$\chi^2$ test results of the 2011 season from least squares of goodness of fit and independence of runs score and allowed. \label{independence}}
\end{figure}

A more in depth discussion of the justification behind the tests can be found in \cite{Mil}.

\section{Future Work and Conclusions}

While a one game improvement in prediction is very promising, as our formula requires us to fit the runs scored and allowed distributions we explored simplifications. We tried to simplify the formula, even giving up some accuracy, in order to devise a formula that could be easily implemented using just a team's runs scored and allowed (and the variance of each of these) in order to determine the team's winning percentage. Unfortunately, the weight parameters $c_1$ and $c_1'$ plays too much of a factor; in 2011, the mean of the parameter $c_1$ is 0.21 with a standard deviation of 0.39 (and a median of 0.21). With such large fluctuations in the weight parameters from team to team, the task of finding a simpler formula was almost impossible, as creating a uniform formula that every team could use was not feasible when two of the key parameters were so volatile. Taking this into account, we tried fixing the $\gamma$, $c_1$, and $c_1'$ parameters, allowing for us to just solve a quartic involving the first and second moments to find the other parameters ($\alpha_{{\rm RS}_1}$, $\alpha_{{\rm RS}_2}$, $\alpha_{{\rm RA}_1}$, and $\alpha_{{\rm RA}_2}$). However, while we were able to solve for the other parameters, plugging these values of the parameters gave us a significantly worse prediction of teams' win-loss percentage compared to linear combination of Weibulls and \url{baseball-reference.com}'s Pythagorean Win-Loss statistic. One of the great attractions of James' Pythagorean formula is its ease of use; we hope to return to other simplifications and approximations in a later paper. Our hope is to find a linearization or approximation of our main result, similar to how Dayaratna and Miller \cite{DaMil1} showed the linear predictor of Jones and Tappin \cite{JT} follows from a linearization of the Pythagorean formula.

To summarize our results, using a linear combination of Weibulls rather than a single Weibull increases the prediction accuracy of a team's W/L percentage. More specifically, we saw that the single Weibull's predictions for a team's wins were on average 4.22 games off (with a standard deviation of 3.03), while the linear combination of Weibull's predictions for a team's wins were from 2004-2012 were on average 3.11 games off (with a standard deviation of 2.33), producing about a $25\%$ increase in prediction accuracy. We also performed $\chi^2$ goodness of fit tests for the linear combination of Weibulls and tested the statistical independence of runs scored and allowed (a necessary assumption), and see that in fact the linear combination of Weibulls with properly estimated parameters obtained from least squares analysis closely maps the observed runs and that runs scored and allowed are in fact statistically independent. In addition, when compared against \url{baseball-reference.com}'s Pythagorean Win-Loss statistic, the linear combination of Weibulls does .06 of a game better in the years from 1979 to 2013, but this improvement cannot be considered statistically significant. However, in more recent years, it is worth noting that it does appear that the linear combination of Weibulls is doing better than \url{baseball-reference.com}'s Pythagorean Win-Loss statistic.

\appendix

\section{Moments of Weibulls}\label{sec:AppMoment}

For the runs scored data, we have $c_2 = 1-c_1$ and the density equals
\begin{equation}
c_1\frac{\gamma}{\alpha_{{\rm RS}_1}} \left(\frac{x-\beta}{\alpha_{{\rm RS}_1}}\right)^{\gamma-1} e^{-(\frac{x-\beta}{\alpha_{{\rm RS}_1}})^{\gamma}}+(1-c_1)\frac{\gamma}{\alpha_{{\rm RS}_2}} \left(\frac{x-\beta}{\alpha_{{\rm RS}_2}}\right)^{\gamma-1} e^{-(\frac{x-\beta}{\alpha_{{\rm RS}_2}})^{\gamma}}.
\end{equation}
From \cite{Mur}, and using the fact that the two Weibulls in the linear combination are independent, we obtained the following moments:
\begin{align}
\textnormal{First Moment}  &\ =\ c_1(\alpha_{{\rm RS}_1}\Gamma(1+\gamma^{-1})+\beta)+(1-c_1)(\alpha_{{\rm RS}_2}\Gamma(1+\gamma^{-1})+\beta) \nonumber \\
\textnormal{Second Moment}  & \ =\  c_1^2\left[ \alpha_{{\rm RS}_1}^2\left( \Gamma(\frac{2}{\gamma}+1)-(\Gamma(\frac{1}{\gamma}+1))^2\right)\right] \nonumber \\ & \indent + (1-c_1)^2\left[ \alpha_{{\rm RS}_2}^2\left( \Gamma(\frac{2}{\gamma}+1)-(\Gamma(\frac{1}{\gamma}+1))^2\right)\right] \nonumber \\
\textnormal{Third Moment} & \ =\  (c_1^3+(1-c_1)^3)\ast \frac{g_3-3g_1g_2+2g_1^3}{(g_2-g_1^2)^{3/2}}\nonumber \\
\textnormal{Fourth Moment} & \ =\  (c_1^4+(1-c_1)^4)\ast \frac{g_4-4g_1g_3+6g_2g_1^2-3g_1^4}{(g_2-g_1^2)^2} \nonumber\\
& \indent + 6\ast c_1^2(1-c_1)^2\ast \left[\alpha_{{\rm RS}_1}^2\alpha_{{\rm RS}_2}^2 \left(g_2-g_1^2\right)^2 \right],
\end{align} where $g_i=\Gamma(1+\frac{i}{\gamma})$, and $\Gamma(x)$ is the Gamma function defined by $\Gamma(x)=\int_0^\infty e^{-u}u^{x-1}du$. We next used $R$ to find the observed moments for teams' runs scored from 2007. The code is available in \cite{Luo}.


text:
\begin{document}

\begin{center}
\Large\sc
Hurwitz stabilisers of some short redundant Artin systems for the braid
group $\br_3$\\[1.1cm]
\end{center}
\normalsize
{\bf \noindent
Michael\ L\"onne},\rm Institut f\"ur Mathematik, Am
Welfengarten 1, 30167 Hannover, Germany (e-mail: loenne@math.uni-hannover.de)

\begin{abstract}
We investigate the Hurwitz action of the braid group $\br_n$ on the $n$-fold
Cartesian product $\br_3^n$ and determine some stabilisers of its Artin systems.

Our algebraic result is complemented by a geometric study of families of plane
polynomial coverings of degree $3$. Together they lead to characterisations of
the set of {\it paths realised by degenerations} of the polynomials as defined
by Donaldson \cite{d}.
\end{abstract}

\section*{introduction}

The braid group $\br_n$ on $n$ strands has first been considered by A.\ Hurwitz
\cite{hu} and E.\ Artin \cite{ar}. Since then it has lead a prolific life in
such
diverse areas of mathematics as topology, combinatorics and algebraic
geometry.\\
In his original approach A. Hurwitz investigates the action of $\br_n$ on the
$n$-fold Cartesian product of a symmetric group. This action can formally be
defined on $G^n$ for any group $G$ and is determined by the action of the
standard generators
$$
\s_i(g_1,...,g_n)=(g_1,...,g_{i-1},g_ig_{i+1}g_i\inv,g_i,g_{i+2},...,g_n)
$$

Braid monodromy factorisations up to Hurwitz equivalence have proved
to be at the heart of the study of complex surfaces and symplectic
four-manifolds, cf.\ \cite{mo,mt,ak,kt}.

On the other hand people looked at stabilisers of Coxeter systems: Given a
symmetric matrix $M$ of positive integers $m_{ij}$ there is the associated
Coxeter group
$$
C=C_M:=\langle s_1,...,s_n\,|\,s_i^2=(s_is_j)^{m_{ij}}=1\rangle.
$$
The Hurwitz stabiliser $S_\cfami$ of the distinguished element
$\cfami=(s_1,...,s_n)\in C^n$ is shown to be
$$
E_M:=\left\langle
e_{ij}:=\s_{j-1}\cdots\s_{i+1}\s_i^{m_{ij}}\s_{i+1}\inv\cdots\s_{j-1}\inv,
1\leq i<j\leq n\right\rangle
$$
in several important cases, \cite{bw,klui,cw,do}.

And there is always a geometric flavour to these results; Birman and Wajnryb
\cite{bw} applied their results to get a presentation for the mapping class
group of surfaces with at most one boundary component and the stabilisers of the
classical Coxeter systems for the finite Coxeter groups
investigated by Catanese/Wajnryb (type $A$) and D\"orner (type $ADE$) had been
shown to be the fundamental groups of the bifurcation complements of the
corresponding simple singularities, cf.\ \cite{lo}.
\\

On the geometric side our research was initiated by the article \cite{d} in
'Mathematics: frontiers and perspectives'.

Given a plane polynomial $g_1=g_1(x,y)$ the projection to the $x$-axis defines
a function $f_1$ on the plane curve $C_1$ defined by $g_1$.
Suppose now $C_1$ is smooth and $f_1$ simply branched, i.e.\ $f_1$ is a Morse
function with singular values $b_0,...,b_n\in\CC$, which are called the branch
points.
Then Donaldson asks for the set of paths which can be realised by deformations
of $g_1$, cf. \cite[ p.56]{d}.
We call such paths {\it vanishing arcs}, since we have the following
characterisation:

An embedded path $\cg$ with endpoints $b_0,b_1$ and disjoint from the branch
set otherwise is a vanishing arc, if there is a smooth family $g_t,t\in[0,1]$
of polynomials and a homotopy $H=H(s,t)$ such that
\begin{enumerate}
\item
the curves $C_t$ are smooth and $f_t$ simply branched for $t>0$,
\item
the curve $C_0$ is smooth except for a single ordinary double point $P$ and $f_0$
is simply branched on the smooth locus and injective on its critical points,
\item
$H$ is a homotopy from $\cg$ to the constant map to $f_0(P)$ such that $H_t$
meets the branch set of $f_t$ in $H_t(\{0,1\})$.
\end{enumerate}
So a vanishing arc can be contracted to point relative to the branch locus.
\\

We first review results on Coxeter systems and tailor them to our needs. In
subsequent sections we investigate stabilisers of Artin systems
$(a_1,...,a_n)\in A^n$ which are Artin analogues of Coxeter systems and
associated to Artin groups
$$
A = A_{M}:=\langle
a_1,...,a_n\,|\,\underbrace{a_ia_ja_i...}_{m_{ij}\text{
factors}}=
\underbrace{a_ja_ia_j...}_{m_{ij}\text{ factors}}\rangle.
$$
To handle degenerations of plane polynomials we choose the set up of polynomial
covers introduced by Hansen \cite{ha} and define the notion of bifurcation
braid monodromy.
The paper closes with a characterisation of sets of vanishing arcs in the
spirit of \cite{d}.
\\

Please note that while we do the computations for all the Hurwitz actions in
the proofs, we only state a lot of not so obvious braid relations which should
be checked by hand or using a symbolic computation package in case of doubt.

We use $\s^\b$ and $(\s)\b$ to denote the action of a braid $\b$ on a braid
$\s$ by conjugation on the right.

\section*{stabilisers of Coxeter systems}

We start with the Hurwitz action on Cartesian powers of the symmetric group
$\sfami_3$ on three elements. The Coxeter presentation for $\sfami_3$ is
generated by two transpositions $s$ and $t$:
$$
\sfami_3\,\cong\,C:=\langle s,t\,|\,s^2=t^2=(st)^3=1\rangle.
$$
Any non-constant $n$-tuple $(s_1,...,s_n)$ with $s_i\in\{s,t\}$ is not only an
element in $C^n$ but a Coxeter system with Coxeter group $C$ and matrix
$$
M:=\big(m_{ij}\big)_{1\leq i,j\leq n},\text{ with } m_{ij}=\left\{
\begin{array}{lll}
3 & \text{if} & s_i\neq s_j\\
1 & \text{if} & s_i=s_j
\end{array}\right.
$$

For some of these Coxeter systems the stabiliser group can be found with
results
of Birman and Wajnryb.

\begin{lemma}
\labell{bw-gen}
The stabiliser group of the Coxeter systems $(s,t,t,...,t)$ and
$(s,s,t,...,t)$ can
be given by
\begin{eqnarray*}
S_C & = & \langle
\s_1^3,\s_2,...,\s_{n-1},
(\s_4)\s_3\s_2\s_1^2\s_2\s_3^2\s_2\s_1 \rangle,\\[4mm]
\text{respectively}\\[3mm]
S_C & = & \langle
\s_1,\s_2^3,\s_3,...,\s_{n-1},
(\s_5)\s_4\s_3\s_2^2\s_3\s_4^2\s_3\s_2,
(\s_3)\s_2\inv\s_1^{-2}\s_2^2\s_1 \rangle.\\
\end{eqnarray*}
\end{lemma}

\proof
The first claim is actually proved in \cite{bw}. The argument can then be
extended to cover the other case, cf.\ \cite{klui}.
\qed

Starting from this lemma we can get the stabiliser groups of the alternating
Coxeter systems $\cfami_n:=(s,t,s,t,...)\in C^n$ with matrix
$$
\label{matrix}
M_n:=\big(m_{ij}\big)_{1\leq i,j\leq n},\text{ with } m_{ij}=\left\{
\begin{array}{lll}
3 & \text{if} & i\not\equiv j \mod 2\\
1 & \text{if} & i\equiv j \mod 2
\end{array}\right.
$$

\begin{lemma}
\labell{conj-elm}
The stabiliser groups $S_{\cfami_n}$ of the alternating Coxeter systems
of length $n$ are conjugated to the stabiliser group of a
Coxeter system $(s,t,t,...,t)$ or $(s,s,t,...,t)$ of equal length. For
small $n$
we have:
$$
\begin{array}{llrcl}
S_{(s,t,s)} & = & \s_1\inv &S_{(s,t,t)}& \s_1\\
S_{(s,t,s,t)} & = & \s_2\inv\s_1 &S_{(s,t,t,t)}& \s_1\inv\s_2\\
S_{(s,t,s,t,s)} & = & \s_3\inv\s_2\s_1\inv &S_{(s,t,t,t,t)}& \s_1\s_2\inv\s_3\\
S_{(s,t,s,t,s,t)} & = & \s_4\inv\s_3\s_2\inv &S_{(s,s,t,t,t,t)}&
\s_2\s_3\inv\s_4\\
\end{array}
$$
\end{lemma}

\proof
The general claim is immediate from the observation that the alternating
system is in the same $\br_n$ orbit as at least one of the reference systems.\\
For the second claim we compute the action of the conjugating element on the
alternating system of two transpositions $s,t$ in the permutation group of $3$
elements, with $r$ denoting the third transposition.
$$
\begin{array}{rcr@{}l}
\s_1  (\un{s,t},s) &
= & & (r,s,s)\\[3mm]
\s_1\inv\s_2  (s,\un{t,s},t) &
= & \s_1\inv & (\un{s,r},t,t)\\
& = & & (r,t,t,t)\\[3mm]
\s_1\s_2\inv\s_3  (s,t,\un{s,t},s) &
= & \s_1\s_2\inv & (s,\un{t,r},s,s)\\
& = & \s_1 & (\un{s,r},s,s,s)\\
& = & & (t,s,s,s,s)\\[3mm]
\s_2\s_3\inv\s_4  (s,t,s,\un{t,s},t) &
= & \s_2\s_3\inv & (s,t,\un{s,r},t,t)\\
& = & \s_2 & (s,\un{t,r},t,t,t)\\
& = & & (s,s,t,t,t,t)\\[3mm]
\end{array}
$$
Surely the stabiliser does not depend on the choice of two non-commuting
transpositions and hence we are done.
\qed

Next we show that the $E_n:=E_{M_n}$ is a set of stabilisers and that its set
of generators can easily be reduced.

\begin{lemma}
\labell{E-gen}
The group $E_n$ is generated by $\s_1^3,\s_{i+1}\s_i\s_{i+1}\inv$,
$i=1,...,n-2$ and is contained in the stabiliser subgroup $S_{\cfami_n}$ of the
alternating Coxeter system $\cfami_n$.
\end{lemma}

\proof
First note that the given generators are just $e_{12}, e_{i,i+2}$,
$i=1,...,n-2$. For the remaining $e_{ij}$ we have the following relations:
$$
\begin{array}{rcll}
e_{ij}&=&(e_{i,i+2})e\inv_{i+2,i+4}e\inv_{i+4,i+6}...e\inv_{j-2,j}, &
i\equiv j\mod 2,\\
e_{ij}&=&(e_{12})e_{13}e_{24}...
e_{i-1,i+1}e\inv_{i+1,i+3}e\inv_{i+3,i+5}...e\inv_{j-2,j}, & i\not\equiv
j\mod 2.
\end{array}
$$
For the second claim it thus suffices to check that
$$
\begin{array}{r@{}lcr@{}lcr@{}lcl}
\s_1^3 & (s,t) & = & \s_1^2 & (r,s) & = & \s_1 & (t,r) & = & (s,t),\\
\s_2\s_1\s_2\inv & (s,\un{t,s}) & = & \s_2\s_1 & (\un{s,s},r) & = & \s_2 &
(s,\un{s,r}) & = & (s,t,s).
\end{array}
$$
\pagebreak
\qed[0mm]

In fact we can choose two additional stabilisers to generate the stabiliser
group $S_{\cfami_n}$. Our choice
$\tau_1=(\s_1)\s_2\s_3\inv\s_4,\tau_2=(\s_2)\s_3\s_4\inv\s_5$ is motivated by
their action on Artin systems, cf.\ lemma \ref{tauArtin}.

\begin{lemma}
\labell{alt-gen}
The stabiliser groups $S_{\cfami_n}$ of the alternating Coxeter systems
of lengths $n=3,4,5,6$ can be given as follows:
$$
\begin{array}{lclllcl}
S_{(s,t,s)} & = & E_3,  & &
S_{(s,t,s,t)} & = & E_4,\\
S_{(s,t,s,t,s)} & = & \langle E_5,\tau_1\rangle,  & &
S_{(s,t,s,t,s,t)} & = & \langle
E_6,\tau_1,\tau_2\rangle.
\end{array}
$$
\end{lemma}

\proof
To show that $\tau_1,\tau_2$ belong to the stabilisers of $\cfami_5,\cfami_6$
resp.\, it obviously suffices to show $\tau_1\in S_{\cfami_5}$:
$$
\begin{array}{crl}
& \s_4\inv\s_3\s_2\inv\s_1\s_2\s_3\inv\s_4 & (s,t,s,\un{t,s})\\
= & \s_4\inv\s_3\s_2\inv\s_1\s_2\s_3\inv & (s,t,\un{s,r},t)\\
= & \s_4\inv\s_3\s_2\inv\s_1\s_2 & (s,\un{t,r},t,t)\\
= & \s_4\inv\s_3\s_2\inv\s_1 & (\un{s,s},t,t,t)\\
= & \s_4\inv\s_3\s_2\inv & (s,\un{s,t},t,t)\\
= & \s_4\inv\s_3 & (s,t,\un{r,t},t)\\
= & \s_4\inv & (s,t,s,\un{r,t})\\
= & & (s,t,s,t,s)
\end{array}
$$
So the given groups are shown to be stabilising.

To prove the inverse implication we note that the stabiliser groups are
generated by the following elements which are obtained from generators of lemma
\ref{bw-gen} using conjugation as provided by lemma \ref{conj-elm}:
$$
\begin{array}{ll}
n=3: &
 \s_1^3,(\s_2)\s_1 ,\\
n=4: &
 (\s_1^3)\s_2,(\s_2)\s_1\inv\s_2,(\s_3)\s_2 ,\\
n=5: &
 (\s_1^3)\s_2\inv\s_3,(\s_2)\s_1\s_2\inv\s_3,
(\s_3)\s_2\inv\s_3,(\s_4)\s_3,\\
& \hspace{7mm}(\s_4)\s_3\s_2\s_1^2\s_2\s_3^2\s_2\s_1\s_1\s_2\inv\s_3,\\
n=6: &
 (\s_1)\s_2\s_3\inv\s_4,(\s_2^3)\s_3\inv\s_4,(\s_3)\s_2\s_3\inv\s_4,
(\s_4)\s_3\inv\s_4,(\s_5)\s_4,\\
& \hspace{7mm}(\s_5)\s_4\s_3\s_2^2\s_3\s_4^2\s_3\s_2\s_2\s_3\inv \s_4,
(\s_3)\s_2\inv\s_1^{-2}\s_2^2\s_1\s_2\s_3\inv\s_4.
\end{array}
$$
The claim now follow from the fact that all these elements can be expressed by
elements of the given groups:
$$
\begin{array}{lrlrlrl}
n=3: && \s_1^3 &&(\s_2)\s_1\\
& \hspace{1mm}= & e_{12}& \hspace{6mm}= & e_{13}\\[1mm]
n=4: && (\s_1^3)\s_2 && (\s_2)\s_1\inv\s_2 && (\s_3)\s_2\\
& = & (e_{12})e_{23}\inv e_{13}\inv & = & (e_{13})e_{23}
& \hspace{6mm}= & e_{24}\\[2mm]
n=5: && (\s_1^3)\s_2\inv\s_3 && (\s_2)\s_1\s_2\inv\s_3 && (\s_3)\s_2\inv \s_3\\
& = & (e_{34})e_{13}\inv & = & (e_{24})e_{34}e_{13}\inv & = & (e_{24})e_{34}
\\[1mm]
&& (\s_4)\s_3 &&
\multicolumn{3}{l}{(\s_4)\s_3\s_2\s_1^2\s_2\s_3^2\s_2\s_1\s_1\s_2\inv\s_3}\\
& = & e_{35} & = &
\multicolumn{3}{l}{(\tau_1)e_{35}e_{45}e_{34}e_{15}\inv}\\[2mm]
n=6: && (\s_2^3)\s_3\inv\s_4 && (\s_3)\s_2\s_3\inv\s_4 && (\s_2)\s_3\inv\s_4\\
& = & (e_{45})e_{24}\inv & = & (e_{35})e_{45}e_{24}\inv & = & (e_{35})e_{45}
\\[1mm]
&& (\s_5)\s_4 &&
\multicolumn{3}{l}{(\s_5)\s_4\s_3\s_2^2\s_3\s_4^2\s_3\s_2\s_2\s_3\inv\s_4}\\
& = & e_{46} & = &
\multicolumn{3}{l}{(\tau_2)e_{46}e_{56}e_{45}e_{26}\inv}\\[1mm]
&& (\s_1)\s_2\s_3\inv\s_4 &&
\multicolumn{3}{l}{(\s_3)\s_2\inv\s_1^{-2}\s_2^2\s_1\s_2\s_3\inv\s_4}\\
& = & \tau_1 & = &
\multicolumn{3}{l}{e_{13}}
\end{array}
$$
\pagebreak
\qed

While the claim of the preceding lemma holds true for $n>6$ the next result
does not and is therefore special to $n=5,6$.

\begin{lemma}
\labell{normal}
In case $n=5,6$ the group $E_n$ is normal in the stabiliser group
$S_{\cfami_n}$.
\end{lemma}

\proof
It suffices to
show that conjugation by $\tau_1$ resp.\ $\tau_1,\tau_2$ maps generators of
$E_n$ to $E_n$:
$$
\begin{array}{llcllcl}
n=5,6: &
(e_{12})\tau_1\inv & = & (e_{45})e_{24}\inv, &
(e_{13})\tau_1\inv & = & (e_{13})e_{12}\inv e_{24} e_{45}e_{24}\inv\\
& (e_{24})\tau_1\inv & = & e_{24} &
(e_{35})\tau_1\inv & = & (e_{35})e_{15}e_{12}\inv e_{15}\inv
e_{12}\inv e_{45}\\[2mm]
n=6: &
(e_{23})\tau_2\inv & = & (e_{56})e_{35}\inv, &
(e_{24})\tau_2\inv & = & (e_{24})e_{23}\inv e_{35} e_{56}e_{35}\inv\\
& (e_{35})\tau_2\inv & = & e_{35} &
(e_{46})\tau_2\inv & = & (e_{46})e_{26}e_{23}\inv e_{26}\inv
e_{23}\inv e_{56}\\
&(e_{12})\tau_2\inv & = & (e_{56})e_{35}\inv e_{13}\inv, &
(e_{56})\tau_1\inv & = & (e_{12})e_{15}\inv e_{12}\inv e_{45}e_{46}

\end{array}
$$
\qed

\section*{stabilisers for redundant Artin systems}

In analogy with the previous section we start with a standard presentation for
the braid group on three strands generated by two braids $a$ and $b$:
$$
A:=\langle a,b\,|\,aba=bab\rangle.
$$
As we may expect the alternating Artin systems $\afami_n:=(a,b,a,b,...)$ of
length $n\geq2$ have matrix $M_n$ and Artin group $A$.

\begin{lemma}
\labell{Hact}
Let $A$ act by elementwise conjugation on the $n$-fold
Cartesian product $A^n$, then the stabiliser subgroup of the alternating Artin
system $\afami_n$ has trivial intersection with the subgroup $H=\langle
a^2,b^2\rangle$ if $n\geq2$.
\end{lemma}

\proof
By the positive solution of the Tits conjecture \cite{cp}, $H$ is freely
generated by
$a^2$ and
$b^2$. On the other hand if $n\geq2$ any element in the stabiliser
must actually
belong to the center of $A\cong\br_3$. But the intersection of the center of
$A$ with $H$
must be contained in the center of $H$ which is trivial since $H$ is free.
\qed

\begin{lemma}
\labell{tauArtin}
The braids $\tau_1,\tau_2$ act on elements of the $H$-orbit of $(a,b,a,b,a,b)$
by overall conjugation with $b^{-2}$ resp.\ $a^{-2}$.
\end{lemma}

\proof
Since overall conjugation commutes with the braid action it suffices to
prove the claim
for the action of $\tau_1,\tau_2$ on $(a,b,a,b,a,b)$,
even one of these cases
suffices by symmetry:
$$
\begin{array}{crl}
& \s_4\inv\s_3\s_2\inv\s_1\s_2\s_3\inv\s_4 & (a ,b,a ,\un{b,a} ,b)\\[2mm]
= & \s_4\inv\s_3\s_2\inv\s_1\s_2\s_3\inv & (a ,b,\un{a ,ba
b\inv},b,b)\\[2mm]
= & \s_4\inv\s_3\s_2\inv\s_1\s_2 & (a ,b,ba b\inv,ba \inv
b\inv
a ba b\inv,b,b)\\[2mm]
= & \s_4\inv\s_3\s_2\inv\s_1\s_2 & (a \un{,b,ba
b\inv},b,b,b)\\[2mm]
= & \s_4\inv\s_3\s_2\inv\s_1 & (\un{a ,bba b\inv
b\inv},b,b,b,b)\\[2mm]
= & \s_4\inv\s_3\s_2\inv & (a bba b\inv b\inv a \inv,\un{a
,b},b,b,b)\\[2mm]
= & \s_4\inv\s_3 & (a bba b\inv b\inv a
\inv,b,\un{b\inv a b,b},b,b)\\[2mm]
= & \s_4\inv & (a bba b\inv b\inv a \inv,b,b\inv
a bbb\inv a \inv b,\un{b\inv
a b,b},b)\\[2mm]
= & & (a bba b\inv b\inv a \inv,b,b\inv a ba
\inv b,b,b\inv b\inv a bb,b)
\end{array}
$$
And the last line equals $(b^{-2}a b^2,b ,b^{-2}a
b^2,b,b^{-2}a
b^2,b)$ since
$$
a bba b\inv b\inv a \inv=a ba \inv ba
b\inv a \inv=b\inv a bbb\inv a \inv b=
b\inv a ba \inv b=b\inv b\inv a bb.
$$
{}\qed

\begin{cor}
\label{corr}
The braid $\tau_1$ acts trivially on alternating Coxeter systems but
non-trivially on
alternating Artin systems for $n\geq5$.
\end{cor}

\begin{prop}
\labell{semi-prod}
The groups $S_{\cfami_5},S_{\cfami_6}$ are semi direct products of their normal
subgroups $E_5$ resp.\ $E_6$ and a free subgroup freely
generated by $\tau_1$ resp.\ $\tau_1,\tau_2$.
\end{prop}

\proof
The subgroup generated by $\tau_1$ resp.\ $\tau_1,\tau_2$ acts freely on
the $H$-orbit
of $(a ,b,a ,b,a )$ resp.\ $(a ,b,a ,b,a ,b)$,
and therefore is a free subgroup. The
claim is then immediate from the normality of the group $E_n$ in $S_{\cfami_n}$
for $n=5,6$, cf.\ lemma \ref{normal}.
\qed

\begin{thm}
Let $S_{\afami_n},S_{\cfami_n}$ be the stabilisers of the Artin resp. Coxeter
system associated with the
$n\times n$ matrix
$$
M_n:=(m_{ij}),\,1\leq i,j\leq n,\text{ with } m_{ij}=\left\{
\begin{array}{lll}
3 & \text{if} & i\not\equiv j \mod 2\\
1 & \text{if} & i\equiv j \mod 2
\end{array}\right.
$$
and denote by $E_n$ the subgroup of the braid group $\br_n$ generated by
elements
\linebreak
$e_{ij}:=\s_{j-1}\cdots\s_{i+1}\s_i^{m_{ij}}\s_{i+1}\inv\cdots\s_{j-1}\inv,
1\leq i<j\leq n$,
then the following relations hold:
$$
\begin{array}{cccccll}
E_n & = & S_{\afami_n} & = & S_{\cfami_n} & \,\,\,\text{ if} & n=2,3,4,\\
E_n & = & S_{\afami_n} & < & S_{\cfami_n} & \,\,\,\text{ if} & n=5,6,\\
E_n & \leq & S_{\afami_n} & < & S_{\cfami_n} & \,\,\,\text{ if} & n\geq7.
\end{array}
$$
\end{thm}

\proof
By straightforward calculation the inclusion $E_n\leq S_{\afami_n}$ is shown,
whereas the inclusion $S_{\afami_n}\leq S_{\cfami_n}$ is obvious.
Equality $E_n=S_{\cfami_n}$ for $n=2,3,4$ is shown in lemma \ref{alt-gen}.
Strict inclusion $S_{\afami_n}< S_{\cfami_n}$ for $n\geq5$ follows from
corollary
\ref{corr}. In case $n=5,6$ finally each braid in $S_{\cfami_n}$ can be
written as a product
$\tau e$ with
$\tau\in\langle\tau_1,\tau_2\rangle$ and $e\in E_n$ by prop.\
\ref{semi-prod}. Hence
it is immediate by lemma \ref{Hact} that $E_n$ is the stabiliser group
$S_{\afami_n}$ in these cases.
\qed

\section*{conjugacy classes of simple braids}

The aim of this section is to exhibit the set of simple braids in $E_n$ as a
single $E_n$ conjugation class for $n\leq6$, where a braid is called simple if
it is isotopic to a half twist associated to a path connecting two punctures.

We extend the conjugation action and its exponential notation to sets, so the
conjugation orbit of $e_{13}$ in $E_n$ is denoted by
$e_{12}^{E_n}:=\{e_{12}\}^{E_n}$.

\begin{lemma}
\labell{Scases}
For $n\leq6$ there is a set of braids $T_{B/S}$ such that $\br_n=T_{B/S}\cdot
S_{\cfami_n}$ and for all $\cg\in{T}_{B/S}$:
\begin{eqnarray*}
e_{13}^\cg\in E_n & \Rightarrow & e_{13}^\cg\in e_{13}^{E_n}.
\end{eqnarray*}
\end{lemma}

\proof
We defined $S_{\cfami_n}$ to be a stabiliser for the Hurwitz action of $\br_n$ on
the finite set $\sfami_3^n$, hence the stabiliser is of finite index. Our
strategy is to construct $T$ as a Schreier left transversal for the
$S_{\cfami_n}$-cosets.

First note that the cosets are in bijection to the elements of the
$\br_n$-orbit of the Coxeter system. Moreover we may assume that the Schreier
transversal contains only positive braids since $\s_i\inv$ and $\s_i^2$ act
the same way on all elements of $\sfami_3^n$.

Such transversal $T_{B/S}$ for $E_6$ can be found, e.g.\ by a short symbolic
computation, containing $240$ elements. Since $E_n= S_\afami$ we extract a
list of all elements $\cg$ such that $e_{13}^\cg$ stabilises $\afami_6$.

For such elements we get in fact $18$ elements $e_{13}^\cg$, which can be shown
to be conjugated in $E_n$ to elements $e_{i,i+2}$ and hence to $e_{13}$:
$$
\begin{array}{lll}
(e_{13})\s_3\s_1=(e_{24})e_{13}\inv &
(e_{13})\s_1\s_2=(e_{13})e_{12}\inv \\
(e_{13})\s_3\s_2\s_2=(e_{13})e_{23}e_{24}\inv &
(e_{13})\s_2\s_2\s_3=(e_{13})e_{23}e_{24} \\
(e_{13})\s_3\s_3\s_1\s_2=(e_{24})e_{12}e_{34} &
(e_{13})\s_3\s_4\s_1\s_1=(e_{24})e_{34}e_{35}e_{12} \\
(e_{13})\s_3\s_3\s_2\s_1\s_1=(e_{24})e_{12}\inv e_{24}\inv &
(e_{13})\s_3\s_2\s_2\s_3\s_1=(e_{24})e_{34}e_{13}^{-2} \\
(e_{13})\s_2\s_2\s_3\s_3\s_1=(e_{24})e_{23} &
(e_{13})\s_2\s_2\s_3\s_4\s_1=(e_{24})e_{34}e_{35} \\
(e_{13})\s_3\s_3\s_4\s_4\s_2=(e_{12})e_{23}e_{35}\inv e_{45} &
(e_{13})\s_3\s_3\s_4\s_5\s_2=(e_{13})e_{23}e_{35}\inv e_{45}e_{46} \\
(e_{13})\s_3\s_4\s_4\s_5\s_2=(e_{13})e_{23}e_{35}\inv e_{45}e_{46}e_{24} &
(e_{13})\s_3\s_4\s_4\s_2\s_1\s_1=(e_{35})e_{45}e_{13}\inv \\
(e_{13})\s_3\s_4\s_5\s_2\s_1\s_1=(e_{35})e_{45}e_{13}\inv e_{46} &
(e_{13})\s_3\s_4\s_2\s_3\s_3\s_1=(e_{24})e_{34}e_{35}\inv \\
(e_{13})\s_3\s_4\s_5\s_1\s_1\s_2=(e_{13})e_{12}e_{35}\inv e_{45}e_{13}
\inv e_{12}\inv e_{46}\\
(e_{13})\s_3\s_4\s_4\s_5\s_5\s_2\s_3\s_1=(e_{46})e_{56}e_{24}e_{35}e_{56}
\end{array}
$$
\qed

\begin{lemma}
\labell{tcases}
Let $\tau\in S_{\cfami_5}$, resp.\ $S_{\cfami_6}$ be a freely reduced word in
$\tau_1$, resp.\
$\tau_1,\tau_2$, then
$$
e_{13}^\tau\in e_{13}^{E_n}.
$$
\end{lemma}

\proof
For the at most four words of
length one we have:
$$
\begin{array}{lcl}
(e_{13})\tau_1 & = & (e_{13})e\inv_{35}e_{45}e_{35}e_{45}e_{13}e_{12}\inv \\
(e_{13})\tau_1\inv & = & (e_{13})e_{12}\inv e_{23}e_{45}e_{24}\inv\\
(e_{13})\tau_2 & = & e_{13}\\
(e_{13})\tau_2\inv & = & e_{13}
\end{array}
$$
To conclude by induction on the word length we next consider a word $\tau$ which
is the concatenation of a word
$\tau'$ and a letter $\tau''$. By induction hypothesis
$e_{13}^{\tau'}=e_{13}^e$ for some $e\in E_n$ and by the normality of
$E_n$ in $\langle E_n,\tau_1\rangle$, resp.\ $\langle
E_n,\tau_1,\tau_2\rangle$ we can write
$e\tau''=\tau''e'$ for a suitable $e'\in E_n$. Hence
$$
e_{13}^{\tau}=e_{13}^{\tau'\tau''}=e_{13}^{e\tau''}
=e_{13}^{\tau''e'}\in e_{13}^{E_n}.
$$
\qed

\begin{prop}
\labell{conclass}
The intersection of the conjugacy class of half twists in
$\br_n,\,n\leq6$, with $E_n$ coincides with the conjugacy class of $e_{13}$
in $E_n$.
\end{prop}

\proof
One inclusion is obvious. So we pick any $\b\in\br_n$ such that $e_{13}^\b\in
E_n$, which may be factorised as $\b=\cg\tau e$ with $\cg\in T_{B/S}$,
$\tau\in\langle\tau_1,\tau_2\rangle$, $e\in E_n$.
Since $\tau e$ normalises $E_n$, we get $e_{13}^\cg\in E_n$ and
$e_{13}^\cg=e_{13}^{e'}$ for some $e'\in E_n$ by lemma \ref{Scases}.

By lemma \ref{normal} and lemma \ref{tcases} there are $e'',e'''\in E_n$ such
that:
$$
e_{13}^{\b} =e_{13}^{\cg\tau e} =e_{13}^{e'\tau e}
=e_{13}^{\tau e''e} =e_{13}^{e'''e''e} \in e_{13}^{E_n}.
$$
\qed

\section*{bundles and monodromy}

We owe the following exposition of polynomial covers to \cite{cs} and
\cite{ha}.

On a connected topological manifold $X$ a simple Weierstrass polynomial of
degree~$d$ is
a map $f:X\times\CC\to\CC$ given by
$$
f(x,z):=z^d+\sum_{i=1}^d c_i(x)z^{d-i},
$$
with continuous coefficient maps $c_i:X\to\CC$, and with no multiple roots
for any
$x\in X$.
Given such a function $f$ the first coordinate projection map onto $X$ may be
restricted
to the subspace
$$
Y_f:=\{(x,z)\in X\times\CC\,|\,f(x,z)=0\}
$$
defining a $d$-fold cover $\pi_f$ onto $X$, the polynomial cover associated
to $f$, or
to the complement $X\times\CC\setminus Y_f$ defining a fibre bundle over
$X$ with fibre
diffeomorphic to a $d$-punctured disc, the punctured disc bundle associated
to $f$.

A finite unramified cover is called polynomial if it is equivalent to a
polynomial
cover for some simple Weierstrass function as above. Any cover $\pi:Y\to
X$ gives rise to a monodromy homomorphism from $\pi_1(X,x)$ to the symmetric
group $\sfami(\pi\inv(x))$,
which serves for a natural characterisation of polynomial covers:

\begin{prop}[\cite{ha}]
\labell{alg-lift}
An unramified cover of degree $d$ is  polynomial if and only if its monodromy
homomorphism to the symmetric group $\sfami_d$ lifts along the natural
homomorphism $\br_d\to\sfami_d$.
\end{prop}

There is a natural way to get from a Coxeter system of
length~$n$ of a
symmetric group $\sfami_d$ to a finite cover: Given an $n$-punctured disc and
a geometric
basis for its fundamental group, which is a choice of free generators
$\cg_1,...,\cg_n$ such that
\begin{enumerate}
\item
the generator are represented by disjoint paths freely homotopic to positive
loops around single punctures,
\item
the product $\cg_1\cdots\cg_n$ is freely homotopic to the positive boundary of
the disc.
\end{enumerate}

A homomorphism to $\sfami_d$ is obtained by assigning to
these generators the elements of the Coxeter system. The preimage
of any
subgroup isomorphic to $\sfami_{d-1}$ determines subgroups of the fundamental
group in a unique conjugacy class and thus a well defined finite cover of the
punctured disc.\\

The corresponding result associates with an Artin system of length~$n$ for
the braid
group $\br_d$ a $d$-punctured disc bundle, once a geometric basis for the
fundamental
group of an $n$-punctured disc has been chosen. Here the basis elements are
mapped to
the generators of the Artin system, so a homomorphism to the braid group is
obtained.\\
Since the space of monic polynomials is an Eilenberg-MacLane space for the
braid group,
there is a smooth classifying map for this homomorphism. Pulling back the
tautological
simple Weierstrass polynomial we get a simple Weierstrass polynomial on the
$n$-punctured disc, and the associated punctured disc bundle is the one we
aim for.

\section*{families of polynomials on the plane}

We enter now the realm of complex geometry where there is an abundance of
covers and
bundles as defined in the previous section.

\begin{description}
\item[Example 1:]
Given a branched cover $Y\to X$ of a complex manifold $X$, the restriction
to the
branch complement is a finite topological cover. Its monodromy is also
called the
monodromy of the branched cover.

\item[Example 2:]
Given a plane curve $C\in\CC^2$ and a projection $p:\CC^2\to\CC$ such that
$p|_C$ is a
finite branched cover, then restricted to the preimage of the complement of
the branch
locus $p|_C$ is a polynomial cover and $p|_{\CC^2-C}$ is a punctured disc
bundle. The
corresponding structure homomorphism to the braid group is called the braid
monodromy.

\end{description}

The second example leads a straight way to the following generalised notion
of braid
monodromy:

\begin{description}

\item[Definition:]
Given a divisor $D\subset T\times\CC$ such that the map $pr|_D$ induced
from the first
projection $pr=pr_1$ onto $T$ is a finite cover, the restriction of $pr$ to the
intersection of the complement of $D$ and the preimage of the branch
complement is a
punctured disc bundle and its structure homomorphism is called the
(generalised) braid
monodromy of $D$.

\end{description}

In favourable circumstances this notion can be used to assign a braid
monodromy to a
family of polynomials.

\begin{description}

\item[Definition:]
A map $f:T\times\CC^2\to\CC$ is called a family of plane polynomials
admissible with
respect to a projection $p:\CC^2\to\CC$ if
\begin{enumerate}
\item
the restriction $f_t$ to each plane $\{t\}\times\CC^2$ is a polynomial,
\item
the zero divisor $Z_f=f\inv(0)$ and the singular values divisor $V_f$ are
branched covers
for the appropriate maps.
$$
\begin{array}{ccc@{\hspace*{1.5cm}}ccc}
Z_f & \to & T\times\CC^2 & V_f & \to & T\times\CC\\
&& \downarrow &&& \downarrow\\
&& T\times\CC &&& T
\end{array}
$$
\end{enumerate}
In this case the generalised braid monodromy of $V_f\subset T\times\CC$ is
called the
bifurcation braid monodromy of the family.

\end{description}

\begin{lemma}
\labell{a-two}
The bifurcation braid monodromy of the family $p_\l(x,y)=y^3-3\l y+2x$ is
generated by
the cube of the twist on the two singular values.
\end{lemma}

\proof
The divisor of singular values is given by the equation $\l^3=x^2$, hence the
bifurcation braid monodromy is the well known braid monodromy of a generically
projected simple cusp.
\qed

\begin{lemma}
\labell{a-one}
The bifurcation braid monodromy of the family $p_\l(x,y)=y^2-x^2+\l$ is the
full braid
group $\br_2$.
\end{lemma}

\proof
The divisor of singular values is given by the equation $x^2=\l$, hence the
bifurcation
braid monodromy is the well known braid monodromy of a vertical tangency
point on a
smooth double cover.
\qed

\begin{description}

\item[Definition:] The bifurcation braid monodromy group of a plane
polynomial $p_0$
with zero set a simple cover branched at $n$ points by a linear projection
$p:\CC^2\to\CC$ is the subgroup of $\br_n$ generated by the images of the
bifurcation
braid monodromy of all families of plane polynomials containing $p_0$ which are
admissible w.r.t.\ $p$.

\end{description}

\begin{prop}
\labell{deg-two}
The bifurcation braid monodromy of any generic polynomial deformation
equivalent to
$y^2-x^k$ is the full braid group $\br_k$.
\end{prop}

\proof
It suffices to consider the family $y^2-x^k+kx+\l$. Its singular value
divisor is given
by the equation $x^k-kx=\l$ of which the braid monodromy is as claimed.
\qed

Thus having dealt with the easiest cases we now want to investigate
polynomials with
branch degree three, which are in fact intimately related to the alternating
Artin systems considered in the first part of this paper.

\begin{lemma}
\labell{nat-basis}
The polynomial cover and its complement fibration for the polynomial
$y^3-3y+2x^k$ are
associated to the alternating Artin system of length $2k$ for a natural
choice of basis
of the fundamental group of the branch complement.
\end{lemma}

\proof
By straightforward computation the fibre at $x=0$ is $\CC^2$ with punctures at
$y=-\sqrt{3},0,\sqrt{3}$ which is regular, non-regular fibres occur at
$x^{2k}=1$ exactly
and along rays $x=r\zeta, r\in[0,1],\zeta$ primitive with $\zeta^{2k}=1$ the
points
$-\sqrt{3},0$ respectively $0,\sqrt{3}$ get closer and merge finally
according to
$\zeta^k=-1$ resp. $\zeta^k=1$.\\
So the elements of the star shaped basis are assigned alternating the twists
of the
intervals $[-\sqrt{3},0],[0,\sqrt{3}]$ which constitute the generator set
for an
alternating Artin system of length $2k$ generating $\br_3$.
\qed

Number the branch points of the polynomial cover given by the polynomial
$y^3-3y+2x^k$
according to increasing $arg$ starting with $x_1=1$.

Notice that the branch locus for the polynomial cover given by
$y^3-3p(x)y+2q(x)$ is
described by the equation $p^3(x)=q^2(x)$.\\

\begin{lemma}
\labell{S-deg}
The family $y^3-3y+2(x^k-\l),\l\in[-1,1]$ degenerates at $\l=\pm1$ only, branch
points are confined to straight rays, and even resp.\ odd indexed branch
points merge
at zero for $\l\to1$ resp.\ $\l\to -1$.
\end{lemma}

\proof
The branch points solve the equation
$$
\begin{array}{rccl}
(x^k-\l)^2=1 & \Leftrightarrow &
& x^k=\l+1\\
&&\vee & x^k=\l-1.
\end{array}
$$
The claim follows.
\qed

\begin{lemma}
\labell{S-mod}
The families $y^3-3y+2(x^k\pm1-\mu kx), \mu\in\CC$ small, have an
associated branch
locus divisor locally isomorphic to that of the family $y^2-x^k$.
\end{lemma}

\proof
The branch locus is given by the equation
$$
\begin{array}{cl}
&(x^k\pm1-\mu kx)^2=1\\
\Leftrightarrow & (x^k-\mu kx)(x^k-\mu kx\mp2)=0
\end{array}
$$
The corresponding divisor consists for small $\mu$ of a smooth unbranched
part and the
divisor associated to $y^2-x^k$.
\qed

\begin{lemma}
\labell{FF-deg}
The family $y^3-3(1-\l)y+2(x^k-i\l), \l\in[0,1]$ degenerates at $\l=1$
only, all branch
points are on a circle of modulus depending on $\l$, and pairs
$x_\nu,x_{\nu+1}$, where $\nu$ is even,
merge at $k$ distinct points for $\l\to 1$.
\end{lemma}

\proof
The branch points solve the equation
$$
\begin{array}{rccl}
(x^k-i\l)^2=(1-\l)^3 & \Leftrightarrow &
& x^k=i\l\pm\sqrt{(1-\l)^3}
\end{array}
$$
and one may check that $arg(x_\nu)$ is strictly increasing resp.\ decreasing
with $\l\to1$
for odd resp.\ even index.
\qed

\begin{lemma}
\labell{F-branch}
The family $y^3+\mu y+2(x^k-i-\mu),\mu\in\CC$ small, has an associated branch
locus divisor isomorphic to $k$ copies of the branch locus divisor of the
family $y^3-3\l y+2x$ locally at $\mu=0$.
\end{lemma}

\proof
The branch locus is given by the equation
$(x^k-i-\mu)^2=-\mu^3$. Hence up to invertible factors this equation reads at
each root $\a$ of the left hand side: $(x-\a)^2=\mu^3$. Since the number of
roots is $k$ the claim follows.
\qed

\begin{lemma}
\labell{F-deg}
There is a family of small deformations of the polynomial $y^3-2(x^k-i)$
such that
generically all branch points are simple except for a single double point.
\end{lemma}

\proof
At $\l=1$ the polynomial is of the form $y^3+2(x^k-i)$, hence the singular
values for
the projection are at $x^k=i$. On may choose an arbitrary one of these roots
say $\a$ and
define a perturbation $y^3-3\e(x-\a)y+2(x^k-i)$.
The singular values are now the zero locus of
$$
\e(x-\a)^3-(x^k-i)^2=:(x-\a)^2(\e(x-\a)-p_\a^2(x)).
$$
Assume there is another double root $x(\e)$ for the $\e$-family then
\begin{eqnarray}
\label{root}
\e(x(\e)-\a)-p_\a^2(x(\e)) & = & 0\\
\label{droot}
\wedge\hspace*{9mm} \e-2p_\a'(x(\e))p_\a(x(\e)) & = & 0
\end{eqnarray}
but on the other hand with equation (\ref{root}) also its derivative must
vanish:
\begin{eqnarray*}
0 & = &\frac\del{\del\e}\left(\e(x(\e)-\a)-p_\a^2(x(\e))\right)\\
& = & x(\e)-\a+\e x'(\e)-2p(x(\e))p'(x(\e))x'(\e)\\
& \stackrel{(\ref{droot})}{=} & x(\e)-\a+\e x'(\e)-\e x'(\e)\\
& = & x(\e)-\a.
\end{eqnarray*}
contrary to the assumption that $x(\e)$ is a root different from $\a$.
\qed

\begin{lemma}
\labell{F-mod}
There is a family of polynomials which contains in its interior a family
parameterised by $\l\in[0,1]$ such that $\l=0$ yields the polynomial
$y^3-3y+2x^k$ and for $\l\to1$ the family meets its only degeneration for which
all points branch points remain distinct except for the merging pair $x_1,x_2$.
Moreover the branch locus divisor has a cusp over $\l=1$.
\end{lemma}

\proof
We have to combine the families of the preceding lemmas into a family of two
complex parameters with $\a$ chosen to be the root at which $x_1,x_2$ merge.
Then the interval can be mapped to the parameter space in such a way as to
yield the desired properties.

The final claim follows from lemma \ref{F-branch}.
\qed

Our objective is now reached since the corresponding monodromies generate
the group $E_n$.

\begin{prop}
\labell{bif-mono}
The bifurcation braid monodromy group of the plane polynomials
$y^3-3y+2x^k$ contains a
subgroup in the conjugacy class of the standard isotropy group $E_n$.
\end{prop}

\proof
The case $k=1$ is lemma \ref{a-two}. For $k\geq2$ we consider the family
$y^3-3\ell(x)y+2(x^k-\mu)$
with $\ell(x)$ linear.
We compute the braid monodromy with respect to the natural choice of
geometric basis of
lemma \ref{nat-basis} with only a slightest move of the reference point
from the
origin to a point in
the sector defined by the rays of $x_{2k}$ and $x_1$.\\
It suffices to show that the generators $e_{12},e_{\nu,\nu+2}$ of $E_M$ are
contained in
the braid monodromy.
The triple twist $e_{12}$ is obtained by going around a degeneration as
given by the
family of lemma \ref{F-mod}.
The twists $e_{\nu,\nu+2}, \nu$ even resp.\ odd are generators for the full
braid group on
the even resp.\ odd indexed branch points. They are realised by an appropriate
deformation in the family of lemma \ref{S-mod}.
\qed

\section*{vanishing arcs and the Donaldson problem}

The objective of this last section is to shed some light on the Donaldson
problem of characterising the vanishing arcs among all isotopy classes of paths
in the base of a polynomial covering. We first sharpen the necessary criterion
of \cite{d}. Next we give a sufficient criterion in terms of the bifurcation
braid monodromy. Finally -- after a short digression to the situation for
finite coverings -- we make ends meet in favourable cases with the help of our
algebraic results.\\

First we give Donaldson's original definition of admissible paths, which we
call Coxeter admissible to distinguish them from our more restrictive
definition of Artin admissible paths:

\begin{description}

\item[Definition:]
A path in the base of a polynomial covering is called {\it Coxeter admissible},
if the monodromy of the finite cover along simple positive small loops around
the endpoints is one and the same transposition, if compared along the path.

\item[Definition:]
A path is called {\it Artin admissible}, if the braid monodromy assigned to
small simple positive loops around the endpoints is one and the same braid, if
compared along the path.

\item[Remark:] We could as well give a definition in terms of the geometry of
degeneration at the ends of a path. The concurrent pairs of points for the
degeneration to both ends of a Coxeter admissible path coincide. This pair
concurs even along isotopic paths in $\CC$ for the degeneration to both ends of
an Artin admissible path.

\end{description}
\pagebreak

\begin{lemma}
\labell{arc-adm}
A vanishing arc is Coxeter and Artin admissible.
\end{lemma}

\proof
A vanishing arc arises in a smoothing of a single ordinary double point.
So locally the divisor of critical values consists of only smooth
components without vertical tangents except for a single smooth component
locally isomorphic to a double cover branched at a single point.

Hence the vanishing arc is Artin admissible as claimed.
\qed

\begin{thm}
\labell{bif-orb}
The set of vanishing arcs for a plane polynomial $y^3-3y+2x^k$ is the orbit of a
single vanishing arc
under the action of the bifurcation braid monodromy group.
\end{thm}

\proof
Any vanishing arc is obtained by a degeneration of the reference polynomial
along an embedded path in the bifurcation complement with second endpoint a
generic point on the degeneration divisor.
Since versal families of the given polynomials are versal for the plane curve
singularity $y^3+x^k$, the degeneration
divisor is irreducible so we may assume within each class of such a path to have
chosen one which ends at a specified point. Hence a pair of vanishing arcs
defines an
element of the fundamental group by concatenation of the corresponding
paths. The braid associated to this loop maps one of the arcs into the other.
\qed

This result should be seen in contrast to the situation where instead of a
family of polynomial covers we consider (abstract) finite covers. Then the set
of corresponding vanishing arcs is much larger as it is even invariant under
the action of $S_{\cfami_{2k}}$
and coincides -- as remarked in \cite{d} -- with the set of Coxeter admissible
arcs.\\

We can finally characterise vanishing arcs as Artin admissible isotopy classes
in case the $x$-degree of our polynomial is sufficiently small:

\begin{thm}
The set of vanishing arcs for the polynomials $y^3-3y+2x^k$, $k=2,3$ is the
orbit of
the chord between the branch points $x_1,x_3$ under the bifurcation braid
monodromy and coincides with the set of Artin admissible paths.
\end{thm}

\proof
For the family of lemma \ref{S-mod} the chord between $x_1,x_3$ is a vanishing
arc and thm.\ \ref{bif-orb} then implies that its orbit under the bifurcation
braid monodromy group $E_{2k}$ is the set of vanishing arcs. By lemma
\ref{arc-adm} it is a subset of the Artin admissible arcs.

On the other hand if we are given an Artin admissible path then performing a
half twist on it does not change the monodromy of the polynomial covering,
hence it corresponds to a half twist in $S_{\afami_{2k}}$. This group coincides
with $E_{2k}$ in the given cases, so the half twist can actually be given as
$e_{13}^{\,e}$ with some $e\in E_{2k}$ by prop.~\ref{conclass}. We conclude
that the given path is the $e$-translate of our chord and hence is a vanishing
arc.
\qed

\begin{description}

\item[Remark:]
In general an Artin admissible isotopy class gives rise to a half twist
contained in $S_{\afami_{2k}}$ and it is an open question whether
$S_{\afami_{2k}}$ equals $E_{2k}$ and whether its half twists are contained in
a single $E_{2k}$ conjugation class.

\end{description}

\pagebreak

config: arxiv


TITLE: Did they find a new particle with CMP?
QUESTION [2 upvotes]: Apparently there has been a paper in nature claiming the finding of a new particle?
http://www.sciencedaily.com/releases/2012/04/120413160004.htm
Does that mean a new fundamental particle has been found with a special design of "common materials"? How does that relate to the standard model?
I hope someone can clarify the meaning of that announcement.

REPLY [5 votes]: Nope, this is just a quasiparticle excitation, like other quasiparticles in condensed matter physics including spinons, chargons, phonons, and others. If they're right, their particular material allows excitations described by fermionic Majorana spinors but they're just emergent fields out of the normal fields that involve leptons, quarks, and the electromagnetic field.
Moreover, I am not sure whether they produce 3+1-dimensional Majorana spinor fields or just lower-dimensional ones. The latter, especially quasiparticles on a wire i.e. in 1+1 dimensions, are of course an easier task.
No implications for particle physics.

REPLY [1 votes]: This a quasi-particle, living in a specially constructed and fined-tuned solid (InSb nanowire in this case). It is not a fundamental particle of the vacuum we find ourselves in. Condesned matter systems offer very rich possibilities to explore both theoretically and experimentally very exotic (quasi-)particles, e.g., anyons.
So no direct implications for the Standrad model whatsoever.


TITLE: Rational plane curves as images of $\mathbb{P}^1$
QUESTION [4 upvotes]: A rational projective plane curve is a curve $C\subseteq \mathbb{P}^2$ such that there exists a birational map $\mathbb{P}^1\dashrightarrow C$.
My question is whether the following statement is true:

A curve $C\subseteq \mathbb{P}^2$ is rational if and only if there exists a morphism of varieties $f: \mathbb{P}^1\rightarrow \mathbb{P}^2$ with $C = f(\mathbb{P}^1)$.

I've been studying rational curves recently and it seems completely natural to expect this to be the case, but I haven't been able to come up with a proof or counterexamples. The image of $\mathbb{P}^1$ by a nonconstant morphism $\mathbb{P}^1\rightarrow \mathbb{P}^2$ is an irreducible curve, and I was hoping that I could describe every rational plane curve in this way. The only if direction seems to be closely related to the question of whether a birational map $\mathbb{P}^1\dashrightarrow C$ can be extended to a morphism $\mathbb{P}^1\rightarrow C$, but I cannot seem to find a reference for the truth of this statement either.
My primary interest is for when the base field is $\mathbb{C}$, if this can affect the answer. Thanks very much for any help!

REPLY [2 votes]: This is true.
First, a birational morphism between curves $C' \dashrightarrow  C$ always extends to a morphism if $C'$ is smooth and $C$ is projective. So being rational is the same as a surjection $f : \Bbb P^1 \to C$ and you are done in one direction.
Conversely, if there is a non-constant morphism $f : \Bbb P^1 \to \Bbb P^2$ , it is a branched cover of its image, and by the Riemann-Hurwitz formula, the image has genus zero. But any genus zero curve is birational to $\Bbb P^1$.
A remark : $C$ will be in general not isomorphic to $\Bbb P^1$, rather it will be a singular curve and $f : \Bbb P^1 \to C$ will be its normalization.


TITLE: How to find the Cartesian equation of a plane in this example (in details)?
QUESTION [0 upvotes]: I'm solving an A Level paper, and came across this question. Basically, they have given plane $p$ has the equation $(\mathbf r-3\mathbf i)\cdot(2\mathbf i-3\mathbf j+6\mathbf k)=0$. Now, I can see that this means dot product, but I don't understand how to find the Cartesian equation (in the form $ax+by+cz=d$). I have no clue about how to do this and so haven't tried to solve it. According to the mark scheme, the normal of the plane is $2\mathbf i-3\mathbf j+6\mathbf k$ and that is all that's given. Thanks in advance!
Edit: couldn't explain the question properly so here's a screenshot:

Here's the Mark Scheme:

Edit: It seems the question is missing details. However, I have provided everything I could possibly provide, along with my understanding of the question (in the comments). I really don't know what to do. If I somehow understand the question and solve it correctly, I will update this question with my answer.

REPLY [1 votes]: Here is what I think after @Lee's comment. That $\mathbf r$ is a generic point on the line $l$ in the first equation, and a generic point of $p$ in the second one. So to intersect $l$ and $p$ you have to make the vector $\mathbf r$ have the form in the first equation and satisfy the second one:
$$\left\{\begin{array}{c}
\mathbf r=(1,6,-3)+s(1,-2,2) \\
(\mathbf r-(3,0,0))\cdot(2,-3,6)=0
\end{array}\right.$$
I suggest you write $\mathbf r=(x,y,z)$. Hence, you will get the following equations:
$$\left\{\begin{array}{l}
x=1+s \\
y=6-2s \\
z=-3+2s \\
(x-3)2+y(-3)+z6=0
\end{array}\right.$$
Solve for $s$, then substitute the result in $(x,y,z)=(1+s,6-2s,-3+2s)$ and that will be $A$. Let me do it:
\begin{align*}
(1+s-3)2-3(6-2s)+6(2s-3)=0\iff{}&s(2+6+12)+2-6-18-18=0\iff{} \\
{}\iff{}&s=\frac{18+18+6-2}{20}=\frac{40}{20}=2.
\end{align*}
Hence:
$$A=(1+2,6-2\cdot2,-3+2\cdot2)=(3,2,1).$$
Now to address the question of the cartesian equations, rewriting the equation yields:
$$(x-3,y,z)\cdot(2,-3,6)=0\iff 2x-3y+6z=6.$$
Geometrically, $p$ is the plane through $(3,0,0)$ perpendicular to $(2,-3,6)$, because a generic point on the plane, if moved by $-(3,0,0)$, ends up on the orthogonal complement to $(2,-3,6)$, which is what the equation given in the paper tells us. Hence, $(2,-3,6)$ is a normal vector (or at least, parallel to a normal vector) to $p$.
While we're at it, let me do the other points of the question. The acute angle, for example. One trick to do that is to compute the angle between the line and the normal vector, and then take $90$ degrees minus that angle to obtain the desired angle. Call the angle between the line and the plane $\alpha$. It will have to satisfy:
$$\cos\alpha=\frac{(2,-3,6)\cdot(1,-2,2)}{\|(2,-3,6)\|\|(1,-2,2)\|},$$
because the line $l$ is parallel to $(1,-2,2)$ and the plane is orthogonal to $(2,-3,6)$. Note that I am doing exactly the "e.g." product in the mark scheme. So:
$$\cos\alpha=\frac{2+6+12}{\sqrt{4+9+36}\sqrt{1+4+4}}=\frac{20}{\sqrt{49}\sqrt{9}}=\frac{20}{7\cdot3}=\frac{20}{21}.$$
Thus:
$$\alpha=\arccos\frac{20}{21}=.30984464\mathrm{rad}.$$
But our angle was:
$$\beta=\frac\pi2-\alpha=1.260951687\mathrm{rad}=72.24720984\text{˚}.$$
Look at the mark scheme: we got it right.
The last point is to find a line lying in $p$, orthogonal to $l$ and passing through $A$. The best way here is to find two linearly independent vectors orthogonal to $(2,-3,6)$, then find a combination of those two orthogonal to $(1,-2,2)$ and find the line through $A$ parallel to that combination. The fact it is a combination of vectors orthogonal to the normal vector to $p$ makes it a vector in $p$, when moved by $(3,0,0)$, the fact it's orthogonal to the direction of $l$ makes it orthogonal to $l$, and then we make it pass through $A$. One easy vector orthogonal to $(2,-3,6)$ is $(0,2,1)$. Another easy one is $(3,2,0)$. These are evidently independent. So let us try finding $a(0,2,1)+b(3,2,0)$ orthogonal to $(1,-2,2)$, that is:
$$(3b,2b+2a,a)\cdot(1,-2,2)=0\iff 3b-4b-4a+2a=0\iff -b-2a=0\iff b=-2a.$$
For example, with $a=1$ we get $b=-2$. So our combination will be $(3(-2),2(-2)+2\cdot1,1)=(-6,-2,1)$. So our line will be $A+s(-6,-2,1)$, or:
$$(3,2,1)+s(-6,-2,1).$$
Obviously our choice of the combination had some degree of arbitrariness. For example, if we multiply $(-6,-2,1)$ by any real number we will get the same line if we use the multiple instead of it in the line's equation. For example, multiplying by $-1$ gives us exactly the equation in the scheme's last OR.