text:
\begin{document}

\title{Intrinsic and Extrinsic curvatures in Finsler{\it esque} spaces}

 \author{Dawood Kothawala}
 \email{dawood@physics.iitm.ac.in}
 \affiliation{Department of Physics, Indian Institute of Technology Madras, Chennai 600 036}

\date{\today}
\begin{abstract}
\noindent We consider metrics related to each other by functionals of a scalar field $\varphi(x)$ and it's gradient $\bm \nabla \varphi(x)$, and give transformations of some key geometric quantities associated with such metrics. Our analysis provides useful and elegant geometric insights into the roles of {\it conformal} and {\it non-conformal} metric deformations in terms of intrinsic and extrinsic geometry of $\varphi$-foliations. As a special case, we compare {\it conformal} and {\it disformal} transforms to highlight some non-trivial scaling differences. We also study the geometry of {\it equi-geodesic} surfaces formed by points $p$ at constant geodesic distance $\sigma(p,P)$ from a fixed point $P$, and apply our results to a specific disformal geometry based on $\sigma(p,P)$ which was recently shown to arise in the context of spacetime with a minimal length.
\end{abstract}

\maketitle
\vskip 0.5 in
\noindent
\maketitle

\section{Introduction} \label{sec:intro}
Consider a spacetime described by a metric $\bm g$, and let $\varphi(x)$ be a given scalar field. From these variables, one can construct another metric on the same manifold defined by
\begin{eqnarray}
\gm_{ab} &=& \mA g_{ab} - \epsilon \mB t_a t_b
\label{eq:qmetric}
\end{eqnarray}
where $\Omega=\Omega[\varphi]$ and $\mB[\varphi]$ are arbitrary functions of the scalar field $\varphi$, and
\begin{eqnarray}
t_a &=& \frac{\partial_a \varphi}{\sqrt{ \epsilon g^{ij} \nabla_i \varphi \nabla_j \varphi }}
\;\; ; \;\;\;
g^{ab} t_a t_b = \epsilon = \pm 1
\end{eqnarray}

Our aim in this paper would be study the relationship between geometric quantities associated with $\bm \gm$ and $\bm g$.

However, before doing that, let us first give a brief motivation for considering such geometries. Under what circumstances can such geometries arise, and what might be their use in physical theories? Over the past several decades, it has been realized (although not widely appreciated) that the geometry relevant for describing matter in presence of gravitational field could be related to the spacetime geometry in a non-trivial manner, going beyond the well known relation based on {\it conformal transformations} which have otherwise been ubiquitous in such contexts. A very transparent presentation of this argument can be found, for example, in Bekenstein's paper \cite{bek-disformal}. He argued that if one takes {\it Finsler}, rather than {\it Riemannian}, geometry as more fundamental, and imposes some physically motivated constraints, a Finsler geometry can be re-cast in terms of a Riemannian geometry with a metric that is related to the spacetime metric by transformations which are non-conformal. Since the transformation in (\ref{eq:qmetric}) is qualitatively of such form (although differs in detail from the ones in \cite{bek-disformal}), we refer to such geometries as ``Finsler{\it esque}", and shall be working throughout this paper in this restricted
context. A more recent reason for considering such geometries comes from semi-classical and quantum gravity. In particular, it was recently shown \cite{dk-ml} that space(time)s with a minimal length scale might be described by an {\it effective} ``metric" of the form (\ref{eq:qmetric}), with the non-local Synge world-function bi-scalar replacing the field $\varphi(x)$. We discuss this geometry in detail in Section \ref{sec:min-length} below.

{\it Summary}: I outline a method which gives, in a convenient form, the expressions for Ricci scalar associated with the metric $\gm_{ab}$ in terms of geometric quantities associated with the level surfaces $\Sigma$ of $\varphi(x)$. In Section \ref{sec:extrinsic}, I focus on the geometry of $\Sigma$ and find several relationships between intrinsic and extrinsic geometric quantities associated with $\Sigma$ as embedded in $\gm_{ab}$ and $g_{ab}$. In Section \ref{sec:modified-riccis}, I use these relations in the Gauss-Codazzi equation to re-construct the Ricci scalar of $\bm \gm$ (see Eqs.~(\ref{eq:mRs-final}, \ref{eq:JcJd})), and give several specific applications, including a recent result in which a non-local disformal coupling based on Synge world function arises naturally in spacetimes with a minimal length. In Section \ref{sec:riccis2}, I recast the derived expression for the Ricci scalar in a manner which elegantly highlights the (well-known) contribution of the conformal part, and the additional contribution due to the non-conformal part expressed in a transparent, geometric way. Finally, I end with brief concluding remarks in Section \ref{sec:concluding-remarks}.

{\it Notation}: The signature is $(-,+,+, \ldots)$ for Lorentzian spaces. I will also use the convenient notation $D_k=D-k$ which is handy when working in $D$ dimensions.

\section{Intrinsic and Extrinsic geometry} \label{sec:extrinsic}

Given the metric $\gm_{ab}$, the inverse metric is easily found to be
\begin{eqnarray}
\gm {}^{ab} &=& \frac{1}{\mA} \; g^{ab} + \epsilon \l( \frac{\Omega^{-2} \mB}{\mA-\mB} \r) q^a q^b
\end{eqnarray}
where $q^a = g^{ab} t_b$ and $\gm {}^{ai} \gm_{ib} = \delta^a_{\phantom{a}b}$.

To begin with, introduce the vectors
\begin{eqnarray}
T_a &=& \sqrt{\mA-\mB} \; t_a
\nn \\
T^a &=& \gm {}^{ab} T_b
\nn \\
&=& \frac{1}{\sqrt{\mA-\mB}} \; q^a
\end{eqnarray}
which are normalized with respect to $\gm_{ab}$. Of course, any characterization of $\Sigma$ in $\gm_{ab}$ must be based on these vectors. The metric determinants are related by
\begin{eqnarray}
\sqrt{-\gm} = \Biggl\{ \Omega^{(D-2)} \sqrt{\frac{\mA-\mB}{\Omega^{-2}}} \Biggl\} \sqrt{-g}
\end{eqnarray}
which follows from the {\it matrix determinant lemma}:
\begin{equation}
{\mathrm{det}} \l(\rm \bf M + \rm \bf u \rm \bf v^{\mathrm{T}}\r) = \l(\mathrm{det} \; \rm \bf M\r) \times \l( 1 + \rm \bf v^{\mathrm{T}} \rm \bf M^{-1} \rm \bf u \r)                                                                                \end{equation}
where $\rm \bf M$ is an invertible square matrix, and $\rm \bf u, \rm \bf v$ are column vectors (of same dimension as $\rm \bf M$).

\begin{center}
\textbf{The first fundamental form}
\end{center}

Using the above relations, one can immediately deduce the following relation between the induced metrics, or the first fundamental forms, of $\Sigma$ in $\bm \gm$ and $\bm g$.
\begin{eqnarray}
\overset{\star}{h}_{ab} &=& \gm_{ab} - \epsilon T_a T_b
\nn \\
&=& \l( \mA g_{ab} - \epsilon \mB t_a t_b \r) - \epsilon \l( \mA-\mB \r) t_a t_b
\nn \\
&=& \mA \; h_{ab}
\end{eqnarray}
That is, the {\it induced} geometries on $\Sigma$ in any two space(time)s related by Eq.~(\ref{eq:qmetric}) are related by a {\it conformal transformation}.

This observation will enormously simplify the evaluation of Ricci scalar of $\gm_{ab}$ in terms of quantities associated with $g_{ab}$.

\begin{center}
\textbf{The intrinsic Ricci scalar}
\end{center}

The above result implies that intrinsic geometries of $\Sigma$ are conformal to each other, and since the conformal factor is $\mA[\varphi]$ which is constant on $\Sigma$, one immediately obtains
\begin{eqnarray}
\overset{\star}{R}_{\Sigma} = \Omega^{-2} R_{\Sigma}
\\
\nn
\end{eqnarray}
which is a simple rescaling, and, most importantly, independent of $\mB$.

\begin{center}
\textbf{The second fundamental form}
\end{center}

Our next aim would be relate the extrinsic geometries, or the second fundamental forms, of $\Sigma$ in the two metrics. These are defined by
\begin{eqnarray}
K_{ij} &=& \nabla_i t_j - \epsilon a_j t_i
\nn \\
\mK_{ij} &=& \mnabla_i T_j - \epsilon \macc_j T_i
\end{eqnarray}
where $a_j$ and $\macc_j$ are the acceleration vectors associated with vectors $q^i$ and $T^i$ respectively.
\begin{eqnarray}
a_j &=& q^k \nabla_k t_j
\nn \\
\macc_j &=& T^k \mnabla_k T_j
\end{eqnarray}

Since $\mnabla_b T_c = \partial_b T_c - \mGamma {}^a_{\phantom{a}bc} T_k$, we need the relation between Christoffel connections of the two metrics. This is given by
\begin{eqnarray}
\mGamma {}^a_{\phantom{a}bc} = \Gamma {}^a_{\phantom{a}bc} + \frac{1}{2} \gm {}^{am}
\l( - \nabla_m \gm_{bc} + 2 \nabla_{(b} \gm_{c)m} \r)
\label{eq:chr-rel}
\end{eqnarray}
We only need $\mGamma {}^k_{\phantom{k}ij} T_k$, which is relatively straightforward to obtain using the identities proved in Appendix \ref{app:chr}, from which we obtain
\begin{eqnarray}
\mnabla_b T_c &=& \sqrt{\mA-\mB} \; \l( \nabla_b t_c + \frac{\nabla_{\bm q} \mA}{2 \l(\mA-\mB\r)} h_{bc} + \frac{\mB}{\mA-\mB} K_{(bc)}  \r)
\nn
\end{eqnarray}
It immediately follows that
\begin{eqnarray}
\macc_c &=& T^b \mnabla_b T_c
\nn \\
&=& \frac{1}{\sqrt{\mA-\mB}} q^b \mnabla_b T_c
\nn \\
&=& a_c
\end{eqnarray}

Putting everything together, and noticing that hypersurface orthogonality implies $K_{(bc)}=K_{bc}$, we finally obtain
\begin{eqnarray}
\mK_{ab} &=& \mnabla_{a} T_b - \epsilon \macc_b T_a
\nn \\
&=& \frac{1}{\sqrt{\mA-\mB}} \Biggl[ \mA K_{ab} + \frac{1}{2} \l( \nabla_{\bm q} \mA \r) h_{ab} \Biggl]
\end{eqnarray}
and
\begin{eqnarray}
\overset{\star}{K} &=& \gm {}^{ab} \mK_{ab}
\nn \\
&=& \frac{\Omega^{-2}}{\sqrt{\mA-\mB}} \Biggl[ \mA K + \frac{D_1}{2} \nabla_{\bm q} \mA \Biggl]
\end{eqnarray}

\begin{center}
\textbf{Summary}
\end{center}

To summarize, we have derived the following relations between first and second fundamental forms of $\varphi(x)=$constant
\begin{eqnarray}
\overset{\star}{h}_{ab} &=& \mA h_{ab}
\nn \\
\overset{\star}{R}_{\Sigma} &=& \Omega^{-2} R_{\Sigma}
\nn \\
\mK_{ab} &=& \frac{\mA}{\sqrt{\mA-\mB}} \Biggl[ K_{ab} + \l( \nabla_{\bm q} \ln \Omega \r) h_{ab} \Biggl]
\nn \\
\mK &=&  \frac{1}{\sqrt{\mA-\mB}} \Biggl[ K + D_1 \nabla_{\bm q} \ln \Omega \Biggl]
\end{eqnarray}

Of special interest are two cases which are displayed in Table~\ref{tab:conformal-vs-disformal}, corresponding to conformal and disformal transformations.

\section{Re-constructing the Ricci scalar from geometry of $\Sigma$} \label{sec:modified-riccis}

Having the relationship between intrinsic and extrinsic geometrical properties of $\Sigma$, we can now re-construct the point wise Ricci scalar of $\gm_{ab}$ by using the Gauss-Codazzi relation:
\begin{eqnarray}
\mRs = \overset{\star}{R}_{\Sigma} - \epsilon \l( \mK {}^2 + \mK {}_{ab}^2 \r) - 2 \epsilon \mnabla_{\bm T} \mK + 2 \epsilon \mnabla_i \macc^i
\label{eq:modrs1}
\end{eqnarray}
where $\mnabla_{\bm T} \equiv T^i \mnabla_i$ and $\mK {}_{ab}^2 \equiv \gm {}^{ia} \gm {}^{jb} \mK_{ab} \mK_{ij}$.

Our aim is to express all the quantities on the RHS of the above equation in terms of quantities associated with $g_{ab}$, thereby obtaining the transformation of $\mRs$. This is easily done by using the results of the previous sections, which give the required relationships for quantities appearing in RHS above.

At this stage, it is useful to trade off the function $\mB$ in terms of a new function $\alpha$, defined by
\begin{eqnarray}
\mA - \mB = \alpha^{-1}
\label{eq:alphadef}
\end{eqnarray}
The significance of introducing $\alpha$ will become clear as we go along. In particular, the set of metric transformations of the form (\ref{eq:qmetric}) turn to have an extremely
simple composition law in terms of the functions $\l(\mA, \alpha\r)$; see Appendix \ref{app:composition-law}.

Using expressions from previous section, one can derive the following relations after few (long) algebraic steps:
\begin{eqnarray}
\mK {}_{ab}^2 &=& \alpha \l[ K_{ab}^2 + K \nabla_{\bm q} \ln \mA + D_1^2 \l(  \nabla_{\bm q} \ln \Omega \r)^2 \r]
\nn \\
\mK {}^2 &=& \alpha \l[ K^2 + D_1 K \nabla_{\bm q} \ln \mA + D_1^2 \l(  \nabla_{\bm q} \ln \Omega \r)^2 \r]
\nn \\
\mnabla_{\bm T} \mK &=& \alpha \nabla_{\bm q} K + \frac{1}{2} \l( K +  D_1 \nabla_{\bm q} \ln \Omega \r) \nabla_{\bm q} \alpha
\nn \\
\phantom{\mnabla_{\bm T} \mK} && \phantom{\alpha \nabla_{\bm q} K } + D_1 \alpha \nabla_{\bm q} \nabla_{\bm q}  \ln \Omega
\nn \\
\mnabla_i \macc^i &=& \Omega^{-2} \nabla_i a^i
\label{eq:aux-eqs}
\end{eqnarray}
where $\nabla_{\bm q} \equiv q^i \nabla_i$ as above.

We now have everything needed to evaluate the RHS of Eq.~(\ref{eq:modrs1}). Putting everything together, and using the (easily proved) identity
\begin{eqnarray}
\nabla_{\bm q} K &=& q^i \nabla_i \nabla_j q^j
\nn \\
&=& - K_{ab}^2 - R_{ab} q^a q^b + \nabla_i a^i
\end{eqnarray}
we get

\begin{eqnarray}
\mRs = \Omega^{-2} R
\;+\; \epsilon \l( \alpha - \Omega^{-2} \r)  \mathcal{J}_d
\;-\; \epsilon \alpha \, \mathcal{J}_c
\label{eq:mRs-final}
\end{eqnarray}
where
\begin{eqnarray}
\mathcal{J}_c &=&
\epsilon \l[ 2 D_1 \Omega^{-1} \square \Omega + D_1 D_4 \Omega^{-2} (\nabla \Omega)^2 \r]
\nn
\\
\vspace{.2cm}
&& \hspace{.55cm} \; + \l( K + D_1 \nabla_{\bm q} \ln \Omega \r) \times \nabla_{\bm q} \ln \alpha \Omega^2
\nn \\
\nn \\
\mathcal{J}_d &=&
2 R_{ab} q^a q^b + K_{ab}^2 -K^2 - 2 \nabla_i a^i
\nn \\
&=& \epsilon \l( R - R_{\Sigma} - 2 \nabla_i a^i \r)
\label{eq:JcJd}
\end{eqnarray}

which is the required expression.

The crucial role played by the {\it non-conformal} term in the metric Eq.~(\ref{eq:qmetric}), characterized by $\mB \neq 0$, is immediately obvious from the above relation. In particular, $ \mathcal{J}_d$ in the above expression is multiplied by $\alpha-\Omega^{-2} = \alpha \Omega^{-2} \mB$ (see Eq.~(\ref{eq:alphadef})). For conformally related metrics, $\mB=0$, and therefore this term does not play any role in conformal transformations. This observation is crucial, since it couples the non-conformal part of the metric $\gm_{ab}$ to the object $\mathcal{J}_d$ which has a very special structure!
\\
\subsection{Applications and Analyses}

\subsubsection{Raychaudhuri equation}

The first application we give is based on the third relation in Eqs.~(\ref{eq:aux-eqs}).

Consider the congruence of integral curves of $q^i$, which need not necessarily be geodesics. Then, the expansion associated with these congruences is $\theta=K$, and it's rate of change along the curves given by $\dot \theta=\nabla_{\bm q} K$. The corresponding quantities in metric $\gm_{ab}$ are given by $\overset{\star}{\theta}=\mK$ and $\overset{\star}{\dot \theta}=\mnabla_{\bm T} \mK $, and their relationship, translated to variables more familiar from the Raychaudhuri equation, is given by
\begin{eqnarray}
\overset{\star}{\l(\frac{\DM \theta}{\DM \lambda}\r) } &=& \alpha \frac{\DM \theta}{\DM \lambda} +  \frac{1}{2} \l( \theta +  D_1 \frac{\DM \ln \Omega}{\DM \lambda} \r) \frac{\DM \alpha}{\DM \lambda} + D_1 \alpha \frac{\DM^2 \ln \Omega}{\DM \lambda^2}
\nn \\
\end{eqnarray}
where we have replaced $\nabla_{\bm q}$ on the RHS with $\DM/\DM \lambda$ - the derivative along the curve. (The special case of conformal transformation corresponds to $\alpha=1/\mA$.)

The above relation gives the modification to the Raychaudhuri equation, and hence should be useful in studying the focussing and de-focussing behavior of geodesics in Finsler-Riemann spaces. If the origin of the scalar field $\varphi(x)$ is due to some fundamental physics associated with quantum gravitational effects, such a modification would be key to understand the effect of quantum gravity on spacetime singularities.

In this context, it is also worth pointing out the relation between the symmetric, traceless part of $K_{ab}$, or the so-called shear-tensor, which happens to be much simpler that the relation between $K_{ab}$ themselves. Using again Eqs.~(\ref{eq:aux-eqs}), we get
\begin{eqnarray}
\overset{\star}{\sigma}_{ab} &=& \mK_{ab} - (1/D_1) \mK \; \overset{\star}{h}_{ab}
\nn \\
&=& \sqrt{\alpha} \; \mA \sigma_{ab}
\end{eqnarray}
and hence
\begin{eqnarray}
\overset{\star}{\sigma} {}^2_{ab} &=& \alpha \; \sigma^2_{ab}
\end{eqnarray}
It is particularly obvious from the above that the conformal ($\alpha=1/\mA$) and disformal ($\alpha=\mA$) cases correspond to completely different scaling of the shear tensor associated with $\varphi(x)$ foliations.

A more general discussion on the propagation of expansion, shear and vorticity in the context of Finsler-Riemann geometries can be found, for example, in \cite{FR-refs}.

\subsubsection{Conformal transformation}

As mentioned at the end of previous section, for conformal transformations, $\alpha=1/\mA$, and the expression for Ricci scalar reduces to
\begin{eqnarray}
\mRs &=& \Omega^{-2} R - \epsilon \Omega^{-2} \l[\mathcal{J}_c\r]_{\alpha \Omega^2=1}
\nn \\
\nn \\
&=& \Omega^{-2} \l[ R - 2 D_1 \Omega^{-1} \square \Omega - D_1 D_4 \Omega^{-2} (\nabla \Omega)^2 \r]
\end{eqnarray}
since $\ln \alpha \Omega^2 = 0$ in this case, and hence the second term in $\mathcal{J}_c$ in Eq.~(\ref{eq:JcJd}) vanishes.

We have therefore re-derived the well known expression for conformal transformation of the Ricci scalar, in a manner which gives a geometric origin for the $\Omega$ dependent  terms.

\subsubsection{Disformal transformation}

For the special case of disformal transformations

\footnote{We should clarify that we are using the term {\it disformal} here to refer to a special subclass of metrics (\ref{eq:qmetric}), while sometimes all
such metrics

are called disformal. This is just a matter of terminology.}

, $\alpha=\Omega^2$, and we obtain
\begin{eqnarray}
\mRs = \Omega^{-2} R + \epsilon \l( \Omega^2 - \Omega^{-2} \r)  \mathcal{J}_d - \epsilon \Omega^2 \l[\mathcal{J}_c\r]_{\alpha = \Omega^2}
\label{eq:Rsdisformal}
\end{eqnarray}

Since disformal modifications of spacetime geometry play an important role in several studies such as modified gravity, the above expression can provide considerable insight into construction of sensible action for such models.

\subsubsection{Small scale structure of spacetime: \\ Disformal coupling through Synge world function bi-scalar} \label{sec:min-length}

\begin{figure*}
    \centering
    \subfloat[The geometry of {\it equi-geodesic} ``foliation".]{{\includegraphics[width=0.7\textwidth]{geodesic-congruence} }}

    \subfloat[Equi-geodesic surface $\Sigma$ in Minkowski spacetime. $\mathcal{H}$ is the null cone at $P$.]{{\includegraphics[width=0.325\textwidth]{equigeodesicsN} }}
    \caption{Set of events $p$ at constant geodesic distance from a given space(time) event $P$. The shaded region represents the normal convex neighborhood $\mathcal N(P)$
	of $P$, and $p \in \mathcal N(P)$.}
    \label{fig:geodesic-congruence}
\end{figure*}

In a recent work \cite{dk-ml}, it was argued that a spacetime with a minimal length is endowed, under certain conditions, with precisely a disformal structure based on the bi-scalar of geodesic distance, $\sigma(p,P)$, between spacetime events $p$ and $P$, with
\begin{eqnarray}
\mA = 1 + \frac{\lp^2}{\sigma(p,P)^2}
\end{eqnarray}
It is a straight forward exercise to plug the above form for $\Omega$ in Eq.~(\ref{eq:Rsdisformal}) and obtain Eq.~(14) of \cite{dk-tp}.

{\it Equi-geodesic surfaces}:

The key role here is played by the congruence of geodesics emanating from a fixed spacetime event $P$, and the surface comprised of events $p$ lying at constant geodesic interval from $P$, which we call as the {\it equi-geodesic} surface. We give below several properties of the foliation based on such surfaces, and also sketch the derivation of Eq.~(14) in \cite{dk-tp} which was based on the analysis presented here. Let us start with the key geometric quantities associated with $\sigma^2=$constant surfaces with one of the events, say $P$, is fixed; see Fig.\;\ref{fig:geodesic-congruence}.

Since the affinely parametrized tangent vector to the geodesic connecting $P$ to $p$ is the normal to the $\Sigma$, and given by  \cite{poisson-lrr}
\begin{eqnarray}
t_a = \frac{\nabla_a \sigma^2}{2 \sqrt{\epsilon \sigma^2}}
\end{eqnarray}
the extrinsic curvature tensor of $\sigma^2=$constant surface, $\Sigma$, is given by
\begin{eqnarray}
K_{ab} &=& \nabla_a t_b
= \frac{\nabla_a \nabla_b \l( \sigma^2/2 \r) - \epsilon t_a t_b}{\sqrt{\epsilon \sigma^2}}
\end{eqnarray}
This particular foliation, formed out of points which are at a fixed geodesic interval from a given point, has many interesting characteristics, all deriving from the fact that the bi-tensor
$\nabla_a \nabla_b \l( \sigma^2/2 \r) $ has a well know covariant Taylor series expansion at $p$ near $P$ \cite{christenson-worldf}:
\begin{eqnarray}
\nabla_a \nabla_b \l( \frac{1}{2} \sigma^2 \r) &=& g_{ab} - \frac{\lambda^2 }{3} \mathcal{E}_{ab} + \frac{\lambda^3}{12} \nabla_{\bm q} \mathcal{E}_{ab}
\nn \\
&-& \frac{\lambda^4}{60} \l(\nabla^2_{\bm q} \mathcal{E}_{ab} + \frac{4}{3} \mathcal{E}_{ia} \mathcal{E}^i_{\phantom{i}b} \r) + O(\lambda^5)
\nn \\
\end{eqnarray}
where $\nabla_{\bm q} \equiv q^i \nabla_i$, $\mathcal E_{ab} = R_{a m b n} q^m q^n$, and $\lambda=\sqrt{\epsilon \sigma^2}$ is the numerical value of the geodesic distance between $P$ and $p$.

Therefore, we see that the extrinsic geometry of such a equi-geodesic ``foliation" is very special, and completely characterized by the {\it tidal tensor} $\mathcal E_{ab} = R_{a m b n} q^m q^n$. In fact, the intrinsic and extrinsic curvatures can be characterized by a systematic Taylor expansion around $P$, given by
\begin{eqnarray}
K_{ab} &=& \frac{1}{\lambda}h_{ab} - \frac {1} {3} \lambda \mathcal{E}_{ab} + \frac {1} {12} \lambda^2 \nabla_{\bm q} \mathcal{E}_{ab} - \frac {1} {60}\lambda^3 F_{ab} + O(\lambda^4)
\nn \\
\nn \\
K &=& \frac{D_1}{\lambda} - \frac {1} {3} \lambda \mathcal{E} + \frac {1} {12} \lambda^2 \nabla_{\bm q} \mathcal{E} - \frac {1} {60}\lambda^3 F + O(\lambda^4)
\nn \\
\nn \\
R_{\Sigma}

&=& \frac {\epsilon D_1 D_2} {\lambda^2} + R - \frac{2\epsilon(D+1)}{3} \mathcal{E} + O(\lambda)
\end{eqnarray}
where $\mathcal{E}=g^{ab} \mathcal{E}_{ab} = R_{ab} q^a q^b$, $F_{ab} = \nabla_{\bm q}^2 \mathcal{E}_{ab} + (4/3) \mathcal{E}_{ak} \mathcal{E}^k_{\phantom{k}b}$, and $F = F_{ab} g^{ab}$.

The {\it exact} form of the Ricci scalar (in which the above Taylor expansions can be plugged if needed) turns out to be

\begin{eqnarray}
\mRs_P(p) &=& \mA \; R(p) - \l( \mA - \Omega^{-2} \r) \times \l( R_{\Sigma} -  R^{\rm flat}_{\Sigma} \r)
\nn \\
&& \hspace{.1cm} + \;
2 \epsilon (\mA-1) \l(D_{-1}/D_1\r) K^{\rm flat} \times \l( K - K^{\rm flat} \r)
\nn \\
\end{eqnarray}

where $R^{\rm flat}_{\Sigma}={D_1 D_2}/{\sigma^2}$ and $K^{\rm flat}={D_1}/{\sqrt{\epsilon \sigma^2}}$ are the induced and extrinsic curvatures of $\sigma^2=$const. surfaces in {\it flat} spacetime. Note that, in flat space(time), these surfaces are simply hyperboloids, and hence maximally symmetric $(D-1)$ spaces with positive or negative curvature; see Fig.\;\ref{fig:geodesic-congruence}(b).

Let us point out some special features of the above expression, which are absent for a generic $\varphi(x)$, being a consequence of (i) disformal nature of the coupling, (ii) the very specific form of $\mA=1+\lp^2/\sigma^2$ which implies that $\Sigma$ corresponds to $\sigma^2=$ constant surfaces.

1. The first two terms on RHS mimic the relationship between $g_{ab}$ and $\gm_{ab}$.

2. The form of the RHS clearly indicates an interplay between $\mA=1$ and $\bm g \equiv$ Riemann flat: the disformal character of the modification $\mA \neq 1$ only couples to the curvature of the background spacetime, that is: {\it flat space(time) is immune to the disformal modification of the above form}!

3. The geometrical structure of the above expression might hold the key to a generic study of behavior a disformal spacetime near curvature singularities of $\bm g$ in terms of focusing and de-focussing of geodesics.

\begin{widetext}

\newsavebox\Conf
\begin{lrbox}{\Conf}
  \begin{minipage}{0.5\textwidth}
    \begin{align*}
     \gm_{ab} &= \mA g_{ab} \\
     & \\
     \overset{\star}{h}_{ab} &= \mA h_{ab} \\
     & \\
      \mK_{ab} &= \Omega K_{ab} + \l( \nabla_{\bm q} \Omega \r) h_{ab} \\
      & \\
      \mK &= \Omega^{-1} K + D_1 \Omega^{-2} \nabla_{\bm q} \Omega
    \end{align*}
  \end{minipage}
\end{lrbox}

\newsavebox\Disf
\begin{lrbox}{\Disf}
  \begin{minipage}{0.5\textwidth}
    \begin{align*}
    	\gm_{ab} &= \mA g_{ab} - \epsilon \l( \mA - \Omega^{-2} \r) t_a t_b \\
	& \\
          \overset{\star}{h}_{ab} &= \mA h_{ab} \\
          & \\
          \mK_{ab} &= \Omega^3 K_{ab} + \l( \mA \nabla_{\bm q} \Omega \r) h_{ab} \\
          & \\
      \mK &= \Omega K + D_1 \nabla_{\bm q} \Omega
    \end{align*}
  \end{minipage}
\end{lrbox}

\begin{table}[h]
  \begin{center}
    \begin{tabular}{ccc}
    \\ \hline \hline \\
      Conformal transformations & & Disformal transformations \\
      & & \\
      $\mB=0; \;\; \alpha = \Omega^{-2}$ & & $\mB = \mA - \Omega^{-2}; \;\; \alpha=\mA$ \\ \hline \hline
      \usebox{\Conf} & & \usebox{\Disf} \\
        \\ \hline \hline \\
    \end{tabular}
  \end{center}
\caption{Comparison of conformal and disformal transforms of {\it first} and {\it second} fundamental forms.}
  \label{tab:conformal-vs-disformal}
\end{table}

\vspace{1cm}
\end{widetext}

\section{Conformal and non-conformal parts of the  Ricci scalar} \label{sec:riccis2}

Let ${\bf Ric}\l[\bm g\r]$ denote the Ricci scalar associated with metric $\bm g$, and define $\mD = \alpha \Omega^2 - 1$ which measures deviation from the conformal case, for which $\alpha = \Omega^{-2}$ and hence $\mD=0$. Then, recalling that
	\begin{eqnarray}
	R_{\Sigma, \bm g} = {\bf Ric}\l[ \bm g \r] + \epsilon \l( K_{ab}^2 + K^2 \r) + 2 \epsilon \nabla_{\bm q} K - 2 \epsilon \bm \nabla \bm \cdot \bm a
	\nn
	\end{eqnarray}
	is the intrinsic Ricci scalar of the $\Sigma$ foliation embedded in $\bm g$, Eq.~(\ref{eq:mRs-final}) can be cast in the following elegant form:
	\begin{widetext}
	\begin{eqnarray}
	{\bf Ric}\l[\mA \bm g - \epsilon \alpha^{-1} \mD \; \bm t \otimes \bm t\r] \; &=& \;
	 \underbrace{
	 \l( 1 + \mD \r) \; {\bf Ric}\l[\mA \bm g \r]
	 }_{\rm the~conformal~part}
	- \; \;
	\Omega^{-2}
	 \underbrace{\l[\;
	 	 \mD \; \overbrace{\l( R_{\Sigma, \bm g} + 2 \epsilon \bm \nabla \bm \cdot \bm a \r)}^{\rm purely~{\it intrinsic}}
		- \;
		\epsilon \dot \mD \;
		\overbrace{\l( K + D_1 \nabla_{\bm q} \ln \Omega \r)}^{\rm purely~{\it extrinsic}}
	\;\r]}_{{\rm contribution~of~the~} \bm t \otimes \bm t {\rm ~term}
	}

	\nn \\
	\nn \\
	\nn \\

		\; &=& \;
 	 \l( 1 + \mD \r) \; {\bf Ric}\l[\mA \bm g \r]
	- \; \;
	 	 \mD \; \l( R_{\Sigma} + 2 \epsilon \bm \nabla \bm \cdot \bm a \r)_{{\mA \bm h}}
		+ \;
		\epsilon \dot \mD \;
		\Omega^{-1}
			K_{\Sigma, {\mA \bm h}}

	\label{eq:rs-geom-struc}
	\end{eqnarray}
	\end{widetext}
where $\dot \mD = \nabla_{\bm q} \mD$.

We believe that the above expression holds the key to the importance of the non-conformal term in Finsler like spaces. It brings in some important geometric quantities associated with the $\varphi$ foliation, and drastically alters the behavior of the Ricci scalar in a manner which pure conformal transformations can not.

The above fact was drastically brought to focus recently in the context of small scale structure of spacetime in presence of a minimal length \cite{dk-tp}, where the non-conformal part leads to some remarkable cancellations which drastically alters the low energy manifestation of minimal length effects, leaving a finite, $O(1)$, relic. This suggests that the non-conformal transformations, if they arise due to quantum gravitational effects, might have non-trivial effect on physics at all scales.

Besides the above, the geometric approach advocated here should also find immediate application in several other physical contexts; we here mention a few, and also point out related generalizations.

{\bf Relativistic MOND}: The class of metrics we have considered here have been most prominent in the context of Modified Newtonian Dynamics (MOND), where disformal transformations have played an
	important role. The relativistic generalization of MOND, proposed by Bekenstein \cite{bek-mond}, involves a metric $\tilde g_{ab}$ constructed from the background metric $g_{ab}$, a
	normalised timelike vector field $U_{a}$, and a scalar field $\varphi$. The structure of this modification has been extensively studied, specifically in the context of {\it cosmology}.
	The geometric formalism presented here should
	be useful to investigate the characteristics of $\tilde g_{ab}$, in particular, for better understanding of the contribution of non-conformal part of the deformation as well structure of modified action for the theory.
	This is already suggested by the discussion in Section
	\ref{sec:riccis2} above. Doing so would require a generalization of the present analysis to the case when the vector field in question $U_a \neq \nabla_a \varphi$. Such a generalization should be straightforward, and
	we hope to address it in future work. We here simply illustrate the power of our method in a simplified context in cosmology.

	Consider the simplest (yet most studied) case of $k=0$ (flat), $D=4$, FLRW metric $g_{ab}$ in standard co-moving coordinates $(t,x,y,z)$ with scale factor $a(t)$.
	To study deformations which respect the
	background symmetry, we assume that $\Omega$ and $\mB$ are functions of $t$ only. Further, since conformal deformations are already
	well understood, we set $\Omega=1$. In such a case, $\alpha=(1-\mB)^{-1}$, and the
	relevant foliation is the one provided by $t=$constant surfaces. For such a foliation, $K=3 \l( \dot a/a \r)$ and $K_{ab}=(1/3) K h_{ab}$. Referring to Eq. (\ref{eq:rs-geom-struc}), and noting that $R_{\Sigma}=0=\bm a$,
	we immediately get the modified Ricci scalar to be
	\begin{eqnarray}
	{\bf Ric}\l[ g_{ab} + \mB \; \delta^0_a \delta^0_b \r] \; &=& \; \frac{ {\bf Ric}\l[ g_{ab} \r]}{1-\mB} - \frac{3 H(t) \dot \mB}{(1-\mB)^2}
	\end{eqnarray}
	where $H(t)=\dot a/a$ is the Hubble parameter. The non-flat case, $k = \pm 1$, is also easy to obtain by noting that the metric $h_{ab}$ (being maximally symmetric) immediately yields $R_{\Sigma, \bm h}=6k/a^2$, which
	can then be plugged in Eq. (\ref{eq:rs-geom-struc}) along with other terms.

	{\bf Lanczos-Lovelock (LL) lagrangians}:
	These class of lagrangians have played an important role in generalizing and understanding various classical and semi-classical aspects of gravity over the past decade
	\cite{llreview}. Therefore, generalizing the calculation for Ricci scalar $R$ presented here to generic LL lagrangians would provide another most natural extension of the analysis presented here. Not only would this extend
	the study of disformal modifications of gravity to Lanczos-Lovelock models of gravity, but it would open up a completely new avenue of research into semi-classical aspects of conformal and non-conformal deformations of
	LL lagrangians. Let us briefly elaborate on this point.
	The LL lagrangians $\mathcal{L}^{\rm bulk}_m$ of order $m$ are built from $m$ copies of curvature tensor $R_{abcd}$, while the surface term $\mathcal{K}^{\rm surface}_m$
	which makes the variational problem well defined is
	built from sum of products of $^{(D-1)}R_{abcd}$ and $K_{ab}$. Therefore, although finding the bulk term requires the knowledge of full modified Riemann tensor (which remains a formidable task), the surface term $
	\mathcal{K}^{\rm surface}_m$, since it
	involves only induced geometry $h_{ab}$, can be evaluated in a straightforward manner from the expressions given here. Since the surface term has played a very important role in semi-classical gravity, especially black
	hole thermodynamics, it's 	disformal transformation can lead to new insights beyond those obtained from purely conformal transformations. We hope to take up this calculation in future work.

\section{Concluding Remarks} \label{sec:concluding-remarks}

The mathematical expressions derived here, and their geometric properties, should be useful in several contexts in which an object like $\gm_{ab}$ arises, either as a `physical metric' to which matter fields couple \cite{bek-disformal}, or through some effective quantum gravitational model \cite{dk-ml, dk-tp}.
We hope that these expressions and the method of derivation would be applicable to a wider class of problems in which the relevant scalar (or bi-scalar) degree of freedom arises through some physical consideration, and couples non-conformally to the spacetime geometry. Moreover, since the Ricci scalar forms the basis for setting up the standard gravitational action, the expression for $\mRs$ in the form given in Eq.~(\ref{eq:rs-geom-struc}) might provide insight into construction of the action involving $g_{ab}$ and $\varphi(x)$.

{\it Acknowledgements --} The author thanks IUCAA, Pune, where part of this work was done, for kind hospitality.

\appendix

\section{Derivation of Christoffel connection component}
\label{app:chr}

We start with the following relations which are straightforward to establish:
\begin{eqnarray}
q^m \nabla_m \gm_{bc} &=& \l( \nabla_{\bm q} \mA \r) g_{bc} - \epsilon \mB \l( t_b a_c + t_c a_b \r) - \epsilon \l(\nabla_{\bm q} \mB\r) t_b t_c
\nn \\
q^m \nabla_b \gm_{mc} &=& - \mB \nabla_b t_c + \epsilon \l[\nabla_{\bm q} \l(\mA-\mB\r)\r] t_b t_c
\label{eq:tmp1}
\end{eqnarray}
where $\nabla_{\bm q} \equiv q^m \nabla_m$. Using Eq.~(\ref{eq:tmp1}) in Eq.~(\ref{eq:chr-rel}), a few steps of algebra give
\begin{eqnarray}
\mGamma {}^a_{\phantom{a}bc} T_a &=& \sqrt{\mA-\mB} ~ \Gamma^a_{\phantom{a}bc} t_a + \frac{1}{2 \sqrt{\mA-\mB}} ~ \Delta_{bc}
\end{eqnarray}
where
\begin{eqnarray}
\Delta_{bc} &=& - \l( \nabla_{\bm q} \mA \r)  \; h_{bc} + \epsilon \l[\nabla_{\bm q} \l(\mA-\mB\r)\r] t_b t_c - 2 \mB K_{(bc)}
\nn
\end{eqnarray}
\\

\section{Composition law of transformations}
\label{app:composition-law}

As an interesting aside, we note the following composition law for the transformations we are considering in this paper. Using the variables $\l(A=\mA, \alpha\r)$, let us consider the following class of metrics, all defined on the same manifold.
\begin{eqnarray}
g^{(1)}_{ab} &=& A_{10} g^{(0)}_{ab} - \epsilon \l( A_{10} - \alpha^{-1}_{10} \r) t^{(0)}_a t^{(0)}_b
\nn \\
g^{(2)}_{ab} &=& A_{21} g^{(1)}_{ab} - \epsilon \l( A_{21} - \alpha^{-1}_{21} \r) t^{(1)}_a t^{(1)}_b
\end{eqnarray}
\\
Then, noting that $t^{(1)}_a = t^{(0)}_a/\sqrt{\alpha_{10}}$, it is easy to show that
\begin{eqnarray}
g^{(2)}_{ab} &=& \l(A_{21} A_{10}\r) g^{(0)}_{ab} - \epsilon \l[ A_{21} A_{10} - \l( \alpha_{21} \alpha_{10} \r)^{-1} \r] t^{(0)}_a t^{(0)}_b
\nn \\
\end{eqnarray}
which immediately yields the following simple composition law
\begin{eqnarray}
A_{20} &=& A_{21} A_{10}
\nn \\
\alpha_{20} &=& \alpha_{21} \alpha_{10}
\end{eqnarray}
or symbolically
\begin{eqnarray}
{g^{(0)}_{ab}}
\;\; \overset{A_{10}, \alpha_{10}}{\xrightarrow{\hspace*{1cm}}} \;\;
g^{(1)}_{ab}
\;\; \overset{A_{21}, \alpha_{21}}{\xrightarrow{\hspace*{1cm}}} \;\;
g^{(2)}_{ab}
\nn \\
\nn \\
{g^{(0)}_{ab}}
\;\;
\underset{\alpha_{20}=\alpha_{21} \alpha_{10}}{
\overset{A_{20}=A_{21} A_{10}}{\xrightarrow{\hspace*{3.4cm}}}
}
\;\;
g^{(2)}_{ab}
\end{eqnarray}

\widetext


TITLE: Proving the relationship $1 + r \leq \left(1 +\frac{r}{m}\right)^m$ for any $m \geq 1.$
QUESTION [2 upvotes]: I am currently trying to prove the following relationship
$$1 + r \leq \left(1 +\frac{r}{m}\right)^m\quad  \text{for any }m \geq 1.$$
Would you be so kind and provide some hints/solutions to the above?

REPLY [4 votes]: You'll probably want to assume at least $r \ge -m$, otherwise it can be false (e.g. try $m=3$ and $r < -9$).
Let $f(r) = (1+r/m)^m - (1+r)$ for $-m \le r < \infty$.  Then $f(0) = f'(0) = 0$, while
$f''(r) = \frac{m-1}{m} (1+r/m)^{m-2} \ge 0$, so $f(r)$ is convex on this interval.  Therefore $f(r) \ge 0$ there.


text:
{\bf Problem.} Lines $l$ and $k$ are parallel to each other. $m\angle A = 120^\circ$, and $m\angle C = 80^\circ$. What is the number of degrees in $m\angle B$?

[asy]
size(100); real h = 1.2; currentpen = fontsize(10pt);
draw(Label("$l$",Relative(1)),(0,0)--(1,0),E);
draw(Label("$k$",Relative(1)),(0,-h)--(1,-h),E);
draw((0,-h)--h/2*(cos(150*pi/180),sin(150*pi/180)) + (0,-h));
draw(Label("$C$",Relative(1)),(0,0)--h*sqrt(3)/2*(cos(-120*pi/180),sin(-120*pi/180)),W);
label("$A$",(0,0),N); label("$B$",(0,-h),S);
label("$120^\circ$",(0,0),SE); label("$80^\circ$",h*sqrt(3)/2*(cos(-120*pi/180),sin(-120*pi/180))+(0.1,0),E);
[/asy]
{\bf Level.} Level 4
{\bf Type.} Prealgebra
{\bf Solution.} We draw segment $\overline{CD}$ as shown parallel to lines $l$ and $k$.

[asy]
size(200); real h = 1.2; currentpen = fontsize(10pt);
draw(Label("$l$",Relative(1)),(0,0)--(1,0),E);
draw(Label("$k$",Relative(1)),(0,-h)--(1,-h),E);
draw((0,-h)--h/2*(cos(150*pi/180),sin(150*pi/180)) + (0,-h));
draw(Label("$C$",Relative(1)),(0,0)--h*sqrt(3)/2*(cos(-120*pi/180),sin(-120*pi/180)),W);
label("$A$",(0,0),N); label("$B$",(0,-h),S);
label("$120^\circ$",(0,0),SE);

pair D = (h*sqrt(3)/2*(cos(-120*pi/180),sin(-120*pi/180))) + (2,0);
draw(D -- (D-(2,0)));
dot(D);
label("$D$",D,E);
[/asy]

Since $\overline{CD}\parallel l$, we have $\angle ACD = 180^\circ - \angle A = 60^\circ$, so $\angle DCB = \angle ACB - \angle ACD = 80^\circ - 60^\circ = 20^\circ$.  Since $\overline{CD}\parallel k$, we have $\angle B = 180^\circ - \angle DCB = \boxed{160^\circ}$.

set_name: MATH


TITLE: Prove that the maximum number of edges in a graph with no even cycles is floor(3(n-1)/2)
QUESTION [7 upvotes]: The question is in the title. I can see why the bound is sharp (for example, a lot of triangles sharing one common vertex if n is odd, or the same but with one spare edge hanging out if n is even). But I can't prove why the bound is the best possible.
I know that each edge must belong to at most one cycle. I was thinking that you could take any path or odd big odd cycle and turn it into bunch of triangles wherever possible, but I think that might not be valid because a lot of greedy local optimums isn't necessarily the global optimum. So what do I do?
I tried looking online for an answer before posting this and all I could find is this: Link
Is the reasoning faulty there? I couldn’t make sense of it after it talked about decomposing into cycles.
Thanks for any help

REPLY [6 votes]: This is not true if graph is not simple, for example take a single vertex and a loop, there is one edge, but your bound equals zero.
On the other hand, if the graph in question is simple then here are some hints:

WLOG you can assume that your graph is a single connected component.
Take some spanning tree of it, you will have $n-1$ edges there.
You are left with more than $\lfloor \frac{n-1}{2} \rfloor$ edges, but observe that you cannot add any vertices (the spanning tree touches all of them), therefore, any edge you add will produce a cycle (at least one).
No edge can be shared among cycles, as this would create an even cycle (this means that each edge you add will create a cycle, but it mustn't create two or more).
The best you can do is form triangles (the smallest possible odd cycles), but you can't obtain more than $\lfloor\frac{n-1}{2}\rfloor$ (for any of the remaining edges you need two from the spanning tree), which is why you can't add yet more edges.

I hope it helps ;-)


TITLE: Why must $\gamma^\mu$ matrices in Dirac equation be linearly independent?
QUESTION [3 upvotes]: In Ashok Das' Lecture on Quantum Field Theory, pg 20, it was said that the Dirac equation is derived by considering the Einstein relation $$p^\mu p_\mu=m$$
as a matrix relation, namely an $n\times n$ identity matrix multiplying both sides.
It was then said that we can take the square root of $p^\mu p_\mu$ to be $\mathcal{P}:$
$$\mathcal{P}=\gamma^\mu p_\mu$$
where $\gamma^\mu$ are four linearly independent $n\times n $ matrices.
I understand the motivation to introduce the matrices $\gamma^\mu$ to make the Einstein relation a matrix equation. But why is there a need for the $\gamma^\mu$ matrics to be linearly independent?

REPLY [1 votes]: In the Dirac representation,the linear independence means   $$\sum_{i=1}^{4} \alpha_{i} \gamma^{i}=0 \Rightarrow \forall i\;, \alpha_{i}=0 $$
If there is a $ \;\alpha_{i}\neq 0 $ for which $\;\sum_{i=1}^{4} \alpha_{i} \gamma^{i}=0\; $ is that we have indirectly * a dimension which depends on the others, which contradicts the very essence of relativity. (4D 'representation' of the universe).
$*\;\;P=(E/c,\vec{p})=mV=m\frac{dX}{dt_{0}}=\gamma m(c,\frac{\vec{x}}{dt})\;\;,\;\;dt=\gamma dt_{0}$


TITLE: Problem 20 from Putnam and Beyond
QUESTION [1 upvotes]: I encountered into problem from book "Putnam and Beyond" which seemed to me quite interesting. However, After a long meditation I can not solve it.
Let $x_1, x_2, \dots, x_n, y_1, y_2, \dots, y_m$ be positive integers, $n,m>1$. Assume that $x_1+x_2+..+x_n=y_1+y_2+\dots+y_m<mn$. Prove that in the inequality $$x_1+x_2+..+x_n=y_1+y_2+\dots+y_m$$ one can suppress (but not all) terms in such a way that the equality is still satisfied.

I read the solution which is attached above. However, I have couple of questions.
1) WLOG they assume that $x_1\geqslant y_1$. However, I dont know how to work with the inverse inequality $x_1\leqslant y_1$. Can anyone explain how to work with it?
2) They state that $y_1$ should be moved back to the right-hand side? What if $x_1-y_1$ is already suppressed by induction hypothesis?
3) Why in the case $y_1<n$ this argument does not work? I can not even understand what namely does work.
The solution, in general, is written in unclear way.
Would be grateful for clear explaining.

REPLY [1 votes]: For $2)$ if that term is already suppressed you don't need to do any more - you have a sum of $x$s $=$ a sum of $y$s.
For $3)$ if $y_1\lt n$ the remaining sum might be too large to apply the inductive hypothesis.


TITLE: Convergence of $\sum \sin^2 \left(\frac{\pi}{n}\right)$
QUESTION [0 upvotes]: How do I show the series converges?
$$\sum_{n=1}^\infty \sin^2 \left(\frac{\pi}{n}\right)$$
Hint is that use comparison test but not many inequalities come to mind.

REPLY [0 votes]: Use the fact that
$$
\sin x \leq x
$$
for all $x\in \mathbb{R}_+$, along with the fact that $\sin$ is odd:
$$
\lvert \sin x\rvert \leq \lvert x\rvert.
$$
Than, a comparison with the "right" $p$-series will do the trick.


TITLE: Functional Encryption for Inner Product Predicates
QUESTION [1 upvotes]: I want to try to implement a functional encryption scheme proposed in http://eprint.iacr.org/2011/410. The first problem I faced with is a TrapGen algorithm. In the paper theorem 3.1 states that:

Let $q, n, m \in \mathbb Z$ with $q \geq 2$ and $m \geq 6n\log q$. There is a p.p.t. algorithm $\mathrm{TrapGen}(q,n,m)$ that outputs a pair $(\mathbf A, \mathbf S) \in \mathbb Z^{n\times m}_q\times \mathbb Z^{m\times m}$ such that $\mathbf A$ is statistically close to uniform in $\mathbb Z^{n\times m}_q$ and $\mathbf S$ is a basis for $\Lambda _q^\bot (A)$, satisfying $\|\mathbf S\| = O(n\log q)$ and $\|\tilde{\mathbf S} \| = O(\sqrt {n\log q})$ with overwhelming probability in n.

Remark: $\tilde {\mathbf S}$ denotes the Gram-Schmidt orthogonalization of $\mathbf S$ and $\|\mathbf S\| = \max _i\|\mathbf s_i\|$.

I want to propose my own $\mathrm{TrapGen}$ algorithm which runs as follows:

Choose $\mathbf A$ uniformly from $\mathbb Z_q^{n\times m}$
Choose $\mathbf R\in \mathbb Z^{n\times m}$ from $\mathcal U_d(-c, c)$ for some $c>0$.

Let $\operatorname{Id}\colon \mathbb Z_q^{m-n\times m}\to \mathbb Z^{m-n\times m}$ be a natural identity map. Then on input $n, m, q$ algorithm outputs a pair $\left(\mathbf A,\left[(\operatorname{Id}(Ker(\mathbf A)) \|q\mathbf R\right]\right)$, where input values and $c$ should satisfy $\|\mathbf S\| = O(n\log q)$ with overwhelming probability.
Does my algorithm applicable for this scheme? If no, how can I correct it? Thanks in advance for any help!

REPLY [4 votes]: Your algorithm will not work, because simply choosing uniformly random $\mathbf{A}$ over $\mathbb{Z}_q$ makes it computationally hard to find any short nonzero vector in $\Lambda^\perp(\mathbf{A})$ -- much less a full basis of short vectors.
The most efficient and implementation-friendly trapdoor generator is the one from Micciancio and Peikert, "Trapdoors for Lattices: Simpler, Tighter, Faster, Smaller," EUROCRYPT '12, available here: http://www.cc.gatech.edu/~cpeikert/pubs/efftrap.pdf
That paper introduces a smaller and more usable notion of trapdoor, which is not a short basis of $\Lambda^\perp(\mathbf{A})$.  The AFV inner product functional encryption scheme can be adapted to work directly with the new trapdoor notion, but that is beyond the scope of this answer.


TITLE: For a diagonalizable matrix $A$
QUESTION [0 upvotes]: '' For a diagonalizable matrix $A$ the matrices $P$ and $D$ such that $A = PDP^{-1}$

REPLY [1 votes]: Some examples:
\begin{align*}
\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}^{-1} \\
&= \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}^{-1} \\
&= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}^{-1}
\end{align*}
In general, the columns of the matrix $P$ can be chosen to be a basis of eigenvectors for the matrix $A$. There will always be infinitely many such bases, as you can always scale any basis vector by a constant. Moreover, you can swap the order of the basis vectors (i.e. swap the order of the columns), which will swap the order of the eigenvalues on the diagonal of $D$. So, $D$ is not unique either.


TITLE: Show that the intersections of the $G_s$ is normal subgroup of $G$
QUESTION [1 upvotes]: I need to prove that given a  group $G$ acting in a set $S$, the intersection of the stabilizers $G_s$, where $G_s:=\{g\in G: g.s=s\}$ and $s$ varies through all $S$, is a normal subgroup of $G$.
But the thing is to find the right homomorphism, or is there an easier way to do it?, Can you help me to prove this result please? Thank you.

REPLY [1 votes]: One can also do the following: The definition of a group operation implies that the map
$$
\Phi : G \to {\rm Sym}_S, g \mapsto \phi_g
$$
with $\phi_g (s) = g.s$ is a homomorphism, where ${\rm Sym}(S)$ is the group of all bijections of $S$.
It is easy to check that
$$
\bigcap_s G_s = {\rm Ker}(\Phi),
$$
and hence normal.


TITLE: I need help doing a basic geometry problem
QUESTION [0 upvotes]: I think it can be solved using parallels but I have no idea how.

Solve for "X"

REPLY [1 votes]: Following @user130512's comment, set up:
$$
90+90+2x+x+B+(360-B)=720
$$


TITLE: Alternate expression for a Taylor Series
QUESTION [1 upvotes]: The exercise goes as follows: Represent $f(x)=\sin(x)$ as the sum of its Taylor series centered at $\pi/3$.
This is how I solved the exercise:
$$
f(x) = f(\pi/3) +\frac{f'(\pi/3)}{1!}(x-\pi/3)+\frac{f''(\pi/3)}{2!}(x-\pi/3)^2+\frac{f'''(\pi/3)}{3!}(x-\pi/3)^3+\cdots
$$
$$
= \frac{\sqrt{3}}{2}+ \frac{1}{2\cdot 1!} (x-\pi/3)-\frac{\sqrt{3}}{2\cdot 2!}(x-\pi/3)^2 - \frac{1}{2\cdot 3!}(x-\pi/3)^3+\cdots
$$
Rearranging all terms in two groups, the ones with $\sqrt(3)/2$ as a common factor and $1/2$, we have:
$$
\left(\frac{\sqrt{3}}{2}+ -\frac{\sqrt{3}}{2\cdot 2!}(x-\pi/3)^2 +\cdots\right)+\left(\frac{1}{2\cdot 1!} (x-\pi/3)- \frac{1}{2\cdot 3!}(x-\pi/3)^3+\cdots\right)
$$
$$
=\frac{\sqrt{3}}{2}\left(1+ -\frac{1}{ 2!}(x-\pi/3)^2 +\cdots\right)+\frac{1}{2}\left(\frac{1}{1!} (x-\pi/3)- \frac{1}{ 3!}(x-\pi/3)^3+\cdots\right)
$$
$$
\frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{(-1)^n(x-\pi/3)^{2n}}{(2n)!}+\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n(x-\pi/3)^{2n+1}}{(2n+1)!}
$$
It seemed simple enough to find the pattern that led to the answer.
To corroborate that my answer was correct, I typed the problem in WolframAlpha and I found this elegant, succinct and perplexing answer:
$$
\sum_{n\ge 0}\frac{(-1)^n\cos(1/6(\pi +3n\pi)) (x-\pi/3)^n}{n!}
$$
Does anyone know how this alternate answer was obtained?
Thank you very much for your time and consideration.

REPLY [2 votes]: Hint: if you use angle-addition we have
$$
\cos(\pi/6 + n\pi/2) = \cos(\pi/6)\cos(n\pi/2) +\sin(\pi/6)\sin(n\pi/2)
$$Then look at what happens if you take $n$ modulo $4$; can you take it from here?


TITLE: Why are totally antisymmetric tensors more useful than totally symmetric tensors?
QUESTION [3 upvotes]: In an arbitrary number of dimensions, one can naturally define two tensors, Kronecker delta and Levi-Civita epsilon tensor. However, why isn't it advantageous to define some totally symmetric tensor as well? Is there an intuitive reason for that or is it basically related to differential forms?

REPLY [7 votes]: The Kronecker delta and the Levi-Civita tensors are very different kinds of tensors.
The Kronecker delta is, abstractly, a linear map:  the identity map.  That's all it is and all it can be.
The Levi-Civita, however, can be interpreted geometrically: as an oriented volumetric subspace (or four-volume or what-have-you).
What do I mean by "oriented"?  Think of the right-hand rule in 3d.  We say that a right-handed coordinate system is somehow oriented differently from a left-handed one.  In fact, choice of orientation has nothing to do with the arrangement of the axes---you could use a "left-handed" orientation even with right-handed axes.  This merely reflects an arbitrary choice that goes on top of the usual structure of the vector space.
An easier example might to think of a sheet of paper, a flat plane.  Draw a spiral on said sheet of paper.  You can turn the paper over, and the spiral runs the opposite direction now, even though the overall plane might not change.  These are two different orientations of the same planar subspace.
A general term for tensors that represent subspaces is a blade.  Vectors are blades: each vector is associated with a line-like subspace.  Two vectors that aren't parallel can define a 2-blade, representing a planar subspace.  Those vectors define an orientation by their ordering, and swapping the vectors reverses, or negates, the orientation.  This is where the nature of blades being antisymmetric tensors comes into play.
It's important to note that linear combinations of 2-blades, called bivectors, are important geometric objects in relativity.  These combinations may no longer be thought of as representing subspaces--for instance, how would you see the $xy$-plane and the $tz$-plane put together as a subspace?--but they are still of great physical significance.  The EM tensor is a bivector.  The Riemann tensor is a linear map from bivectors to bivectors (those bivectors are often broken down into plades, to talk about the planes in which covariant derivatives are considered, for whether they do or do not commute and in what planes they get distorted or changed to).
The highest dimensional blades in a space are often called pseudoscalars.  You might have heard that the magnetic scalar potential is really a pseudoscalar.  The Levi-Civita is really a pseudoscalar.  The action of a linear map can be characterized by feeding it a pseudoscalar instead; geometrically, this is interpreted as the amount by which a volume is dilated or shrunk.  The volume might also change orientation (change sign).  This is the determinant of a linear map.  Cool, huh?
$k$-blades and $k$-vectors are very important kinds of tensors to understand.  They are antisymmetric to capture the idea that interchanging vectors changes the orientation of the underlying geometric object.  There are no corresponding geometric analogues for symmetric tensors that are symmetric across many indices.  Most of the geometry is covered with blades and multivectors.


TITLE: find all plane of symmetry of hyperbolic paraboloid
QUESTION [0 upvotes]: Hi i need a little help to find all plane of symmetry of the  hyperbolic paraboloid. The equation is $x^2-y^2=z$.  I know that "standard" planes are   $x=0$ , because if i replace $x$ with $-x$ the equation does not change and $y=0$ for the same reason . My question is how can i say if there are more planes  of symmetry of this equation  ?( i mean "crooked" planes (i used a translator for this word)  Is there a way to find all plane of symmetry for an equation? thanks a lot

REPLY [1 votes]: Here is a mostly complete answer which I have thought way too long on. First, use homogeneous coordinates so that we can say the surface has equation $X^TQX=0$, where $X = \begin{bmatrix} x & y & z & 1 \end{bmatrix}^T$ and
$$
    Q = \begin{bmatrix} 2 &  0 &  0 & 0 \\
                        0 & -2 &  0 & 0 \\
                        0 &  0 &  0 & -1 \\
                        0 &  0 & -1 & 0 \end{bmatrix}
$$
A (projective) linear transformation $A$ preserves the surface if for all $X$,
$$
    X^TQX = (AX)^T Q (AX) = X^T (A^TQA) X
$$
which would mean $A^TQA = Q$.  I want to say that $A^TQA=Q$ is necessary as well as sufficient, and I believe one can use the fact that $Q$ is invertible to prove that detail.
Now if $A$ is a reflection, $A^{-1} = A$, so $A^TQA = Q$ is equivalent to $A^TQ = QA$.  What is $A$?  From Wikipedia, if a plane has equation $ax+by+cz+d=0$, where $a^2+b^2+c^2=1$, the transformation of projective space which reflects in the plane has matrix equation $X' = AX$, where
$$
    A = \begin{bmatrix} 1 - 2 a^2  & - 2 a b & - 2 a c & - 2 a d \\ - 2 a b  & 1 - 2 b^2 & - 2 b c & - 2 b d \\ - 2 a c & - 2 b c & 1 - 2c^2 & - 2 c d \\ 0 & 0 & 0 & 1 \end{bmatrix}
$$
So $QA = A^TQ$ is equivalent to
$$
\left[\begin{matrix}2 - 4 a^{2} & - 4 a b & - 4 a c & - 4 a d\\4 a b & 4 b^{2} - 2 & 4 b c & 4 b d\\0 & 0 & 0 & -1\\2 a c & 2 b c & 2 c^{2} - 1 & 2 c d\end{matrix}\right]
= \left[\begin{matrix}2 - 4 a^{2} & 4 a b & 0 & 2 a c\\- 4 a b & 4 b^{2} - 2 & 0 & 2 b c\\- 4 a c & 4 b c & 0 & 2 c^{2} - 1\\- 4 a d & 4 b d & -1 & 2 c d\end{matrix}\right]
$$
These matrices are equal when each pair of corresponding entries are equal.  but the entries below the diagonal are redundant with ones above the diagonal, and the entries along the diagonal are tautologies.  So there are at most six equations to work with:
\begin{align*}
    -4ab &= 4ab \tag{1.2} \\
    -4ac &= 0   \tag{1.3} \\
    -4ad &= 2ac \tag{1.4} \\
     4bc &= 0   \tag{2.3} \\
     4bd &= 2bc \tag{2.4} \\
    2c^2-1 &= -1 \tag{3.4}
\end{align*}
The equation label $(m.n)$ comes from the $(m,n)$th entry of the matrices.  Remember also that $a^2+b^2+c^2 = 1$.
Now to solve this system.  First, by $(3.4)$, $c=0$.  Therefore $a^2+b^2 =1$.  If we square both sides of $(1.4)$ and $(2.4)$ and add left and right sides, we get $4(a^2+b^2)d^2 = (a^2+b^2)c^2$.  Since $a^2+b^2=1$ and $c=0$, we are left with $d=0$ as well.
Since $ab=0$ by $(1.2)$, either $a=0$ or $b=0$.  So it looks like there are four solutions for $(a,b,c,d)$: $(\pm1,0,0,0)$ and $(0,\pm1,0,0,0)$.  But it's more like two, since the first two solutions correspond to the same plane, and generate the same matrix $A$, and same for the last two.  So the unique solutions are:

$(a,b,c,d) = (1,0,0,0)$, which gives the plane $x=0$ and the symmetry $x' = -x$, $y' = y$, $z'=z$.
$(a,b,c,d) = (0,1,0,0)$, which gives the plane $y=0$ and the symmetry $x' =x$, $y'=-y$, $z'=z$.

Exercise. Find the planes of symmetry of the other quadric surfaces (e.g., the hyperboloids, elliptic paraboloids, cones, etc.) using this method.  The only change will be the matrix $Q$.


TITLE: verify R is an equivalence relation
QUESTION [1 upvotes]: let $S=\Bbb R\times \Bbb R$ define a relation $R$ on $S$ by $(a,b)R(c,d)$ iff $a-b=c-d$, verify $R$ is an equivalence relation

I don't think that it is reflexive, because $a-b$ will not always $= b-a$.
For example if $a=1$ and $b= 3$, $1-3=-2$, but $3-1=2$.

REPLY [0 votes]: Any relation $R$ defined by $aRb \iff f(a)\equiv f(b)$, where $f$ is a function and $\equiv$ is an equivalence relation, is itself an equivalence relation.


TITLE: Why can you replace $\sin(az)$ with $\exp(iaz)$ in $\int_{-\infty}^\infty \frac{\sin(ax)}{x(x^2+1)}dx$?
QUESTION [0 upvotes]: I am looking at the exercise VII.5.2 from Gamelin's Complex Analysis:

Show using residue theory that
  $$\int_{-\infty}^\infty \frac{\sin(ax)}{x(x^2+1)}dx = \pi(1-e^{-a}),\quad a>0.$$
  Hint. Replace $\sin(az)$ by $e^{iaz}$, and integrate around the boundary of a half-disk indented at $z=0$.

I was wondering

Why can you replace $\sin(az)$ by $e^{iaz}$?

I tried looking over the section on trig functions but I couldn't find an answer. I also tried thinking about it, but I couldn't think of anything. I know that
$$\sin az = \frac{e^{iaz}-e^{-iaz}}{2i}$$
but I don't see how this relates.

REPLY [2 votes]: You've picked the wrong relation between the sine and the exponential.  You want $\mathrm{e}^{\mathrm{i} a z} = \cos( a z) + \mathrm{i} \sin (a z )$.  Do you see that you then only want to know the imaginary part of the integral with $\mathrm{e}^{\mathrm{i} a z}$?
Edit to respond to question in comment below:
\begin{align}    \int_{-\infty}^\infty \frac{\exp(\mathrm{i} a z)}{z(z^2+1)} \,\mathrm{d}z &= \int_{-\infty}^\infty \frac{\cos( a z) + \mathrm{i} \sin(a z)}{z(z^2+1)} \,\mathrm{d}z  \\
    &= \int_{-\infty}^\infty \frac{\cos( a z)}{z(z^2+1)} \,\mathrm{d}z + \mathrm{i} \int_{-\infty}^\infty \frac{\sin(a z)}{z(z^2+1)} \,\mathrm{d}z \text{.}
\end{align}


text:
\begin{document}

\title{Non-minimal bridge positions of torus knots are stabilized}

\author{Makoto Ozawa}

\address{Department of Natural Sciences, Faculty of Arts and Sciences, Komazawa University, 1-23-1 Komazawa, Setagaya-ku, Tokyo, 154-8525, Japan}

\curraddr{Department of Mathematics and Statistics, The University of Melbourne
Parkville, Victoria 3010, Australia (temporary)}
\email{w3c@komazawa-u.ac.jp, ozawam@unimelb.edu.au (temporary)}

\subjclass[2000]{Primary 57M25; Secondary 57Q35}

\keywords{torus knot, bridge decomposition, bridge position, bridge presentation, stabilization, stabilized, unstabilized, stably equivalent}

\begin{abstract}
We show that any non-minimal bridge decomposition of a torus knot is stabilized and that $n$-bridge decompositions of a torus knot are unique for any integer $n$.
This implies that a knot in a bridge position is a torus knot if and only if there exists a torus containing the knot such that it intersects the bridge sphere in two essential loops.

\end{abstract}

\maketitle

\section{Introduction}

Throughout this paper we work in the piecewise linear category.

Knot theory treats all embeddings of the circle $S^1$ into the 3-sphere $S^3$ up to equivalence relation by orientation preserving homeomorphisms of $S^3$, and it is most fundamental and important problem to determine whether given two knots are equivalent and furthermore to describe how one can be deformed to another.

In \cite{Sc}, Schubert introduced the bridge number of knots, that is, the half of minimal number of intersection of a 2-sphere $S$ with a knot $K$ which cuts the pair $(S^3,K)$ into two trivial tangles $(B_1,T_1)$ and $(B_2,T_2)$.
Since a knot is trivial if and only if it has the bridge number $1$, the bridge number measures some complexity of knots.

Two bridge decompositions $(B_1,T_1)\cup_{S}(B_2,T_2)$ and $(B_1',T_1')\cup_{{S}'}(B_2',T_2')$ of a pair $(S^3,K)$ are said to be {\em equivalent} if there exists an orientation preserving homeomorphism of $S^3$ which sends one to another.
Since we may assume that $S={S}'$, $B_1=B_1'$ and $B_2=B_2'$, this condition can be replaced by the following condition: there exist homeomorphisms $f_1$ of $B_1$ and $f_2$ of $B_2$ such that $f_1(T_1)=T_1'$, $f_2(T_2)=T_2'$ and $f_1|_{S}=f_2|_{S}$.
Therefore, the bridge decomposition has a merit in a sense that it can decompose an orientation preserving homeomorphism of $S^3$ into two homeomorphisms of $B_1$ and $B_2$ if the knot admits a unique $n$-bridge decomposition.

In exchange for this, bridge decompositions of a knot $K$ are not unique generally even if it has the minimal bridge number.
For example, Birman (\cite{B}) and Montesinos (\cite{Mo}) gave an example of composite knots and prime knots respectively which admit at least two equivalence classes of minimal bridge decompositions.
The following theorem seems to be well-known since it can be proved by showing that Reidemeister moves (\cite{AB}, \cite{R}) can be achieved by bridge isotopies and stabilizations (this has been also pointed out by Joel Hass).

\begin{theorem}[well-known]
All bridge decompositions of a knot are stably equivalent.
\end{theorem}

This theorem says that any two bridge decompositions are equivalent after some stabilizations.
(The term ``stabilization'' in \cite{B}, \cite{O1}, etc is exchange for ``perturbation'' in \cite{ST}, \cite{T}, etc. since they studied a generalized bridge decomposition of knots in a 3-manifold and the term``stabilization'' has been already used for Heegaard splittings.)
This result corresponds to Reidemeister-Singer's theorem \cite{R}, \cite{S}: all Heegaard splittings of a closed orientable 3-manifold are stably equivalent.
And also, this is compared to Markov theorem \cite{M}: all closed braid representatives of a knot are stably equivalent.

\begin{figure}[htbp]
	\begin{center}
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.7\linewidth]{stabilization.eps}
	\end{center}
	\caption{Stabilization (Perturbation)}
	\label{stabilization}
\end{figure}

Toward the unknotting problem, Otal (later Hayashi and Shimokawa, the author) showed the following theorem.

\begin{theorem}[\cite{O1}, \cite{HS}, \cite{OZ}]\label{Otal}
Any non-minimal bridge decomposition of the trivial knot is stabilized.
\end{theorem}

This theorem says that any two $n$-bridge decompositions of the trivial knot are equivalent since the trivial knot admits a unique 1-bridge decomposition.
Thus, a knot $K$ with an $n$-bridge decomposition $(B_1,T_1)\cup_{S}(B_2,T_2)$ is trivial if and only if there exists a 2-sphere $F$ containing $K$ such that $F\cap S$ consists of a single loop (\cite{OZ}).
This result corresponds to Waldhausen's theorem \cite{W}: any non-minimal Heegaard splitting of the 3-sphere is stabilized, and by Alexander's theorem \cite{A}, any two genus $n$ Heegaard splittings are equivalent.

Furthermore, Otal (later Scharlemann and Tomova) showed the following theorem.

\begin{theorem}[\cite{O2}, \cite{ST}]\label{2-bridge}
Any non-minimal bridge decomposition of a 2-bridge knot is stabilized.
\end{theorem}

This theorem says that any two $n$-bridge decompositions of a 2-bridge knot are equivalent up to reflection since a 2-bridge knot admits at most two 2-bridge decompositions and they are related by the reflection with respect to the bridge decomposing 2-sphere.
This result corresponds to Bonahon-Otal's theorem \cite{BO}: any non-minimal Heegaard splitting of a lens space is stabilized.

In this paper, we show the following theorem.

\begin{theorem}\label{torus}
Any non-minimal bridge decomposition of a torus knot is stabilized.

\end{theorem}

We remark that $b(K)=\min \{p,q\}$ for a $(p,q)$-torus knot $K$ $(p,q>0)$, where $b(K)$ denotes the bridge number of $K$.
This was proved by Schubert \cite{Sc} and later Schultens \cite{Sch}, the author \cite{OZ}.
Furthermore, $n$-bridge decompositions of a torus knot are unique for any integer $n$ since a torus knot admits a unique minimal bridge decomposition.

\begin{corollary}
A knot $K$ with an $n$-bridge decomposition $(B_1,T_1)\cup_{S} (B_2,T_2)$ is a torus knot if and only if there exists a torus $F$ containing $K$ such that $F\cap S$ consists of two essential loops.
\end{corollary}

It is clear that if a bridge decomposition of a knot is minimal, then it is unstabilized.
Theorem \ref{Otal}, \ref{2-bridge} and \ref{torus} show that the converse holds for the trivial knot, 2-bridge knots and torus knots respectively.

\begin{problem}

Is any non-minimal bridge decomposition of a knot stabilized?
\end{problem}

In this respect, Casson and Gordon \cite{CG} (later Kobayashi \cite{K}, Lustig and Moriah \cite{LM}, Moriah, Schleimer and Sedgwick \cite{MSS}) showed that there exists a 3-manifold which admits unstabilized Heegaard splittings of arbitrarily large genus.
Therefore it is expected that there exists a knot which admits non-minimal unstabilized bridge decompositions.

Theorem \ref{Otal} and \ref{torus} show that for any integer $n$, there exists a unique $n$-bridge decomposition of the trivial knot and a torus knot respectively, and Theorem \ref{2-bridge} shows that for any integer $n$, there exist at most two $n$-bridge decompositions of a 2-bridge knot.
Generally, Coward showed the following theorem.

\begin{theorem}[\cite{C}]
There exist only finitely many bridge decompositions of given bridge numbers for a hyperbolic knot.
\end{theorem}

This result corresponds to Li's settlement \cite{L} of the Waldhausen conjecture: there exist only finitely many Heegaard splittings of given Heegaard genus for a closed, orientable, irreducible and atoroidal 3-manifold.

In despite of this, Jang \cite{J} showed that there exists a 3-bridge link which admits infinitely many 3-bridge decompositions.
And also Sakuma \cite{Sa} (later Morimoto and Sakuma \cite{MS}, Bachman and Derby-Talbot \cite{BD}) showed that there exists a 3-manifold which admits infinitely many minimal genus Heegaard splittings.
We also remark that Jang \cite{J2} classified 3-bridge algebraic links and moreover, for non-Montesinos ones, she also given a classification of 3-bridge decompositions of each link.

\section{Definitions and related results}

\subsection{Three definitions of the bridge number}

Let $K$ be a knot in the 3-sphere $S^3$, and $F$ be a 2-sphere embedded in $S^3$ which intersects $K$ in $2n$ points $(n\ge1)$.
Then by the Alexander's theorem \cite{A}, $F$ decomposes $S^3$ into two 3-balls $B_1$ and $B_2$, and $K$ into two collections $T_1$ and $T_1$ of $n$ arcs.
Then the pair $(B_i,T_i)$ is called an {\em $n$-string tangle}, and we say that the pair $(S^3,K)$, or simply $K$, admits an {\em $n$-string tangle decomposition} $(B_1,T_1)\cup_F (B_2,T_2)$.
We say that a tangle $(B_i,T_i)$ is {\em trivial} if there exists a disk $D$ properly embedded in $B_i$ such that $D\supset T_i$, and that $K$ admits a {\em trivial $n$-string tangle decomposition}, or simply an {\em $n$-bridge decomposition}, $(B_1,T_1)\cup_F (B_2,T_2)$ if both of $(B_1,T_1)$ and $(B_2,T_2)$ are trivial.
The {\em bridge number} $b(K)$ is defined as the minimal number of $n$ such that $K$ admits an $n$-bridge decomposition.

There are at least two other definitions of the bridge number as follows.

First suppose that a knot $K$ is disjoint from $\pm\infty=(0,0,0,\pm 1)\in \Bbb{R}^4$ and let $p:S^3-\{\pm\infty\} \to S$ be a projection by regarding $S^3-\{\pm\infty\}$ as $S\times \Bbb{R}$.
Then we say that $p(K)$ is a {\rm regular projection} if it has only transverse double points, and a {\em regular diagram} $\tilde{K}$ of $K$ is a regular projection with an over/under information on its double points, namely {\em crossings}.
We say that a subarc of $\tilde{K}$ is an {\em over bridge} (resp. {\em under bridge}) if it contains only over crossings (resp. under crossings).
A partition $\tilde{K}=K_1^+\cup K_1^-\cup \cdots \cup K_n^+\cup K_n^-$ is called an {\em $n$-bridge presentation} of $K$ if $K_i^+$ is an over bridge and $K_i^-$ is an under bridge for $i=1,\ldots,n$.
The {\em bridge number} $b(K)$ is defined as the minimal number of $n$ such that $K$ admits an $n$-bridge presentation.

Next let $h:S^3\to \Bbb{R}$ be the standard Morse function, that is, the restriction of $\Bbb{R}^4=\Bbb{R}^3\times \Bbb{R}\to \Bbb{R}$ on $S^3$.
We may assume that $K$ is disjoint from $\pm\infty$ and $K$ has only finitely many critical points with respect to $h$.
We say that $K$ is in an {\em $n$-bridge position} if $K$ has just $n$ maximal points and $n$ minimal points, and all maximal (resp. minimal) values are greater (resp. less) than $0$.
Then the 2-sphere $S=h^{-1}(0)$ is called an {\em $n$-bridge sphere} for $K$.

It is well known that the above three definitions are equivalent.
We do not prove this here, but Scharlemann's survey \cite{S1} is a good reference.

\begin{proposition}\label{fact}
The following are equivalent.
\begin{enumerate}
\item A knot $K$ admits an $n$-bridge decomposition with a bridge sphere $S$.
\item A knot $K$ admits an $n$-bridge presentation with a bridge sphere $S$.
\item A knot $K$ admits an $n$-bridge position with a bridge sphere $S$.
\end{enumerate}

\end{proposition}

\subsection{On the connected sum of knots}

Let $K_i$ $(i=1,2)$ be a knot in $S^3$ which admits an $n_i$-bridge decomposition $(B_i^+,T_i^+)\cup_{F_i}(B_i^-,T_i^-)$.
Pick one point $p_i$ of $K_i\cap F_i$ and take a small regular neighborhood $C_i$ of $p_i$ so that the pair $(C_i,K_i\cap C_i)$ is a trivial 1-string tangle.
Then we obtain an $(n_1+n_2-1)$-bridge decomposition $((B_1^+-\text{int}C_1)\cup(B_2^+-\text{int}C_2), (T_1^+-\text{int}(K_1\cap C_1))\cup (T_2^+-\text{int}(K_2\cap C_2))) \cup_{(F_1-\text{int}(F_1\cap C_1)\cup(F_2-\text{int}(F_2\cap C_2))} ( (B_1^--\text{int}C_1)\cup(B_2^--\text{int}C_2), (T_1^--\text{int}(K_1\cap C_1))\cup (T_2^--\text{int}(K_2\cap C_2)))$ for a connected sum $K_1\# K_2$ of $K_1$ and $K_2$, and we denote it by $(B_1^+,T_1^+)\cup_{F_1}(B_1^-,T_1^-)\# (B_2^+,T_2^+)\cup_{F_2}(B_2^-,T_2^-)$.
Thus, we have $b(K_1\# K_2)\le b(K_1)+b(K_2)-1$.
Schubert (later Schultens) showed the converse inequality: $b(K_1\# K_2)\ge b(K_1)+b(K_2)-1$.
The proof involves the following theorem.

\begin{theorem}[\cite{Sc}, \cite{Sch1}, \cite{D}, \cite{HS2}]
For any $n$-bridge decomposition $(B_1,T_1)\cup_{S} (B_2,T_2)$ of a composite knot $K$, there exists a decomposing sphere $F$ for $K$ such that $F\cap S$ consists of a single loop.
\end{theorem}

This theorem says that any bridge decomposition of a composite knot $K_1\# K_2$ can be decomposed into two bridge decompositions of $K_1$ and $K_2$.
This result corresponds to Haken's theorem \cite{H}: for any Heegaard splitting of a reducible closed orientable 3-manifold, there exists an essential 2-sphere which intersects the Heegaard surface in a sigle loop.
This shows that the Heegaard genus is additive under connected sum.
And also, this is compared to Birman-Menasco's theorem \cite{BM}: for any closed braid representatives of a composite knot $K$, there exists a decomposing sphere $F$ for $K$ such that $F$ intersects the braid axis in two points after some exchange moves.
Since an exchange move does not change the braid index, this shows that the braid index $(-1)$ is additive under connected sum.

\subsection{Stabilization and stably equivalent}
Now we define the {\em stabilization} and {\em stably equivalent} of bridge decompositions as follows.
Let $K_0$ be the trivial knot which admits a 2-bridge decomposition $(B_0^+,T_0^+)\cup_{F_0}(B_0^-,T_0^-)$.
It follows by Theorem \ref{Otal} that there exists a 2-sphere which contains the trivial knot and intersects $F_0$ in a single loop.
The {\em stabilization} of an $n$-bridge decomposition $(B^+,T^+)\cup_{F}(B^-,T^-)$ of a knot $K$ is an operation to obtain a new $(n+1)$-bridge decomposition $(B^+,T^+)\cup_{F}(B^-,T^-)\# (B_0^+,T_0^+)\cup_{F_0}(B_0^-,T_0^-)$ of $K\# K_0$ from $(B^+,T^+)\cup_{F}(B^-,T^-)$, and the resultant bridge decomposition is said to be {\em stabilized}.
See Figure \ref{stabilization}.
We say that a bridge decomposition is {\em unstabilized} if it is not stabilized.
Two bridge decompositions $(B_1^+,T_1^+)\cup_{F_1}(B_1^-,T_1^-)$ and $(B_2^+,T_2^+)\cup_{F_2}(B_2^-,T_2^-)$ are {\em stably equivalent} if they are equivalent after some stabilizations, namely, there exist non-negative integers $m,n$ such that $\displaystyle (B_1^+,T_1^+)\cup_{F_1}(B_1^-,T_1^-)\#^m (B_0^+,T_0^+)\cup_{F_0}(B_0^-,T_0^-)$ and $(B_2^+,T_2^+)\cup_{F_2}(B_2^-,T_2^-)\#^n (B_0^+,T_0^+)\cup_{F_0}(B_0^-,T_0^-)$ are equivalent.

\subsection{Weakly reducible bridge decompositions}

Casson and Gordon \cite{CG2} showed that if a Heegaard splitting of a closed orientable 3-manifold $M$ is weakly reducible, then either it is reducible or $M$ contains an essential surface.
This result was generalized by Hayashi and Shimokawa (later Tomova) to a 3-manifold containing a 1-manifold as follows.
A bridge decomposition $(B^+,T^+)\cup_{F}(B^-,T^-)$ of a knot $K$ is {\em weakly reducible} if there exists a pair of compressing disks $D^{\pm}$ for $F-K$ in $B^{\pm}-T^{\pm}$ such that $D^+\cap D^-=\emptyset$, and otherwise it is {\em strongly irreducible}.
A properly embedded surface $S$ in the exterior $E(K)$ of $K$ is an {\em essential meridional surface} if $\partial S$ consists of meridians of $K$ and $S$ is incompressible and $\partial$-incompressible in $E(K)$.
The following theorem has been reduced to bridge positions of a knot in $S^3$.

\begin{theorem}[\cite{HS2}, \cite{T}]\label{WR}
If a bridge position of a knot is weakly reducible, then either it is stabilized or the knot exterior contains an essential meridional surface.
\end{theorem}

\subsection{Essential Morse bridge position}

Following \cite{OZ}, we define a {\em Morse bridge position} of a pair $(F,K)$ of a closed surface $F$ containing a knot $K$ with respect to $h:S^3\to \Bbb{R}$ as follows.
We divide $S^3$ into three parts $M_+$, $M_0$ and $M_-$ so that $M_{\pm}\cap M_0=h^{-1}(c_{\pm})$ is a level 2-sphere for $c_+>0>c_-$.
We say that a pair $(F,K)$ with a bridge decomposition $(B^+,T^+)\cup_{S}(B^-,T^-)$ is in a {\em Morse bridge position} if

\begin{enumerate}
\item $h|_F$ is a Morse function,
\item $K$ is disjoint from critical points of $F$,
\item all maximal (resp. minimal) points of $F$ are contained in $M_+$ (resp. $M_-$),
\item all saddle points of $F$ are contained in $M_0$,
\item $(M_{\pm},K\cap M_{\pm})$ is homeomorphic to $(B^{\pm},T^{\pm})$ as a pair,
\item $(M_0,K\cap M_0)$ is homeomorphic to $(S,K\cap S)\times I$ as a pair,
\item $S=h^{-1}(0)$.
\end{enumerate}

Let $x$ be a saddle point of $F$ which corresponds to the critical value $t_x\in \Bbb{R}$.
Let $P_x$ be a pair of pants component of $F\cap h^{-1}([t_x-\epsilon, t_x+\epsilon])$ containing $x$ for a sufficiently small positive real number $\epsilon$.
Let $C_x^1$, $C_x^2$ and $C_x^3$ be the boundary components of $P_x$, where we assume that $C_x^1$ and $C_x^2$ (``thigh'' loops) are contained in the same level $h^{-1}(t_x\pm\epsilon)$, and $C_x^3$ (a ``waist'' loop) is contained in the another level $h^{-1}(t_x\mp\epsilon)$.
A saddle point $x$ of $F$ is {\em upper} (resp. {\em lower}) if $C_x^1$ and $C_x^2$ are contained in $h^{-1}(t_x-\epsilon)$ (resp. $h^{-1}(t_x+\epsilon)$).
See Figure \ref{upper/lower}.
A saddle point $x$ of $F$ is {\em essential} if both of the two loops $C_x^1$ and $C_x^2$ are essential in $F$, and it is {\em inessential} if it is not essential.
A Morse bridge position of a pair $(F,K)$ is said to be {\em essential} if $F$ has no inessential saddle point.

\begin{figure}[htbp]
	\begin{center}
	\begin{tabular}{ccc}
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.35\linewidth]{upper.eps}&
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.35\linewidth]{lower.eps}\\
	Upper saddle & Lower saddle
	\end{tabular}
	\end{center}
	\caption{A pair of pants $P_x$ containing a saddle point $x$}
	\label{upper/lower}
\end{figure}

\begin{lemma}[\cite{OZ},\cite{Sch}]\label{essential saddle}
A pair $(F,K)$ of a closed surface $F$ containing a knot $K$ with a bridge decomposition $(B^+,T^+)\cup_{S}(B^-,T^-)$ can be isotoped in its equivalence class so that it is in an essential Morse bridge position.
\end{lemma}

\section{Proof of Theorem \ref{torus}}
\begin{proof}
Let $K$ be a $(p,q)$-torus knot $(2\le p<q)$ in $S^3$ which admits a non-minimal $n$-bridge decomposition $(B^+,T^+)\cup_{S}(B^-,T^-)$ $(p<n)$.
Since $K$ is a torus knot, there exists a torus $F$ containing $K$ such that $F$ separates $S^3$ into two solid tori $W_1$ and $W_2$, where we assume that $W_1$ contains a meridian disk $w_1$ with $|\partial w_1\cap K|=p$ and $W_2$ contains a meridian disk $w_2$ with $|\partial w_2\cap K|=q$.

\begin{lemma}
If a bridge decomposition of a torus knot is weakly reducible, then it is stabilized.
\end{lemma}

\begin{proof}
Since the torus knot exterior does not contain an essential meridional surface \cite{Ts}, this lemma follows Theorem \ref{WR}.
\end{proof}

Hereafter, we assume that $(B^+,T^+)\cup_{S}(B^-,T^-)$ is strongly irreducible.

Proposition \ref{fact} shows that $K$ admits an $n$-bridge position with respect to the hight function $h:S^3\to \Bbb{R}$, where $S=h^{-1}(0)$, $B^{\pm}=h^{-1}([0,\pm\infty))$.
Then we apply Lemma \ref{essential saddle} to the pair $(F,K)$.
Since $F$ is a torus, for each essential saddle point $x$ of $F$, the waist loop $C_x^3$ bounds a disk $D_x$ in $F$ which contains only one maximal or minimal point, and the thigh loops $C_x^1$ and $C_x^2$ are mutually parallel essential loops in $F$.
Since all thigh loops are parallel to either $\partial w_1$ or $\partial w_2$ in $F$, each disk $D_x$ contains at least $p$ maximal (resp. minimal) points of $K$, where $x$ is an upper (resp. lower) saddle point.
Two saddle points $x$ and $y$ of $F$ are said to be {\em adjacent} if one of $C_x^1, C_x^2$ and one of $C_y^1, C_y^2$ cobound an annulus in $F$, say $A_{xy}$, which does not contain any critical point.
We call such an annulus $A_{xy}$ an {\em adjoining annulus} between $x$ and $y$.
With this notation, we have a partition $F=(\bigcup_{x\in X} P_x) \cup (\bigcup_{x\in X} D_x) \cup (\bigcup_{x,y\in X} A_{xy})$ of pairs of pants, disks and adjoining annuli, where $X$ is the set of all saddle points of $F$.

The following lemma says that the torus $F$ is also in a ``bridge position'', that is, all upper saddle points are above all lower saddle points.

\begin{lemma}\label{adjoining}

There exists a level 2-sphere $h^{-1}(t)$ $(c_+>t>c_-)$ in $M_0$ which intersects all adjoining annuli.
\end{lemma}

\begin{proof}
This lemma follows the argument of \cite[Lemma 6]{Sch2} and the assumpution that $(B^+,T^+)\cup_{S}(B^-,T^-)$ is strongly irreducible.
If all upper saddle points are above all lower saddle points, then this lemma holds.
Otherwise, there exists a regular value $t_0\in \Bbb{R}$ between adjacent critical values $t_x>t_y$ which correspond to a lower saddle point $x$ and an upper saddle point $y$ respectively, in other words, $h^{-1}(t_0)$ is a ``thin sphere'' for $F$.
Then $D_x'=D_x\cap h^{-1}([t_0,-\infty))$ and $D_y'=D_y\cap h^{-1}([t_0,+\infty))$ are disjoint disks properly embedded in $h^{-1}([t_0,-\infty))$ and $h^{-1}([t_0,+\infty))$ respectively.
Since both of $D_x'$ and $D_y'$ contain at least $p$ minimal or maximal points respectively, if we isotope $D_x'$ and $D_y'$ slightly, then we have two disks which shows that a bridge decomposition given by $h^{-1}(t_0)$ is weakly reducible.
By the condition (6) of Morse bridge position, it follows that $(B^+,T^+)\cup_{S}(B^-,T^-)$ is also weakly reducible.
This contradicts the assumpution.
\end{proof}

By the condition (6) of Morse bridge position, we may assume that the level 2-sphere $h^{-1}(t)$ given in Lemma \ref{adjoining} is $S=h^{-1}(0)$.
Thus $F\cap B^{\pm}$ consists of annuli each of which contains only one essential saddle point and only one maximal or minimal point.
Hereafter we denote this annulus decomposition of $F$ by $F=A_1\cup A_2\cup \cdots \cup A_n$ in this order on the torus $F$, where $n=|F\cap S|$ and $A_i\subset B^+$ (resp. $A_i\subset B^-$) if $i$ is odd (resp. even).

\subsection{The case that $|F\cap S|\ge 4$}

First we consider the case that $|F\cap S|\ge 4$.
Put $C_i=A_i\cap A_{i+1}$ for $i=1,\ldots,n-1$ and $C_n=A_n\cap A_1$.
Let $A_i'$ be an annulus in $S$ which is cobounded by $\partial A_i$ for $i=1,\ldots,n$.

There are three possibilities for a pair of adjacent annuli $(A_i,A_{i+1})$ $(i=1,\ldots,n-1)$ or $(A_n,A_1)$.

\begin{enumerate}
\item ``{\em Normal}'': Neither turning nor Reeb.
\item ``{\em Turning}'': $\text{int}A_i'\cap \text{int}A_{i+1}'=\emptyset$ $(i=1,\ldots,n-1)$ or $\text{int}A_n'\cap \text{int}A_1'=\emptyset$.
\item ``{\em Reeb}'': $A_i'\subset A_{i+1}'$ or $A_i'\supset A_{i+1}'$ $(i=1,\ldots,n-1)$, or $A_n'\subset A_1'$ or $A_n'\supset A_1'$.
\end{enumerate}

\begin{figure}[htbp]
	\begin{center}
	\begin{tabular}{ccc}
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.35\linewidth]{normal.eps}&
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.32\linewidth]{turning.eps}&
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.25\linewidth]{Reeb.eps}\\
	Normal & Turning & Reeb
	\end{tabular}
	\end{center}
	\caption{Adjacent annuli}
	\label{adjacent}
\end{figure}

\begin{lemma}\label{two cases}
Then one of the following holds.
\begin{enumerate}
\item $C_1,\ldots,C_n$ bound mutually disjoint disks $D_1,\ldots,D_n$ in $S$.
\item There exists a turning pair of adjacent annuli.
\end{enumerate}
\end{lemma}

\begin{proof}
We suppose that there exists no turning pair of adjacent annuli, and show that $C_1,\ldots,C_n$ bound mutually disjoint disks $D_1,\ldots,D_n$ in $S$.

If there exists a Reeb pair of adjacent annuli $(A_i,A_{i+1})$ such that $A_i'\supset A_{i+1}'$ without loss of generality, then $(A_{i+1},A_{i+2})$ is also a Reeb pair of adjacent annuli such that $A_{i+1}'\supset A_{i+2}'$ since $C_{i+1}$ is essential in $A_i'$ and there exists no turning pair of adjacent annuli.
As a consequence of this observation, we obtain an infinite sequence $A_i'\supset A_{i+1}'\supset A_{i+2}'\supset \cdots$, a contradiction.
Hence, all pair of adjacent annuli are normal.

We show by an induction on $k$ $(k=1,\ldots,n-1)$ that $C_n,C_1,\ldots,C_k$ bound mutually disjoint disks $D_n,D_1,\ldots,D_k$ in $S$ whose interiors are disjoint from $A_1\cup \cdots \cup A_k$.
We may assume without loss of generality that $C_n$ is innermost in $S$ among all $C_1,\ldots,C_n$.
First, it is clear that $C_n,C_1$ bound mutually disjoint disks $D_n,D_1$ in $S$ whose interiors are disjoint from $A_1$.
Thus the induction holds on $k=1$.
Next suppose that the induction holds for $k<i$ and does not hold on $k=i$.
Then $C_i$ bounds a disk $D_i$ in $S$ with $D_i\cap D_{i-1}=\emptyset$ such that $D_i$ contains some disks of $D_n,D_1,\ldots,D_{i-3}$.
This implies that there exists a ``long Reeb''$A_j\cup A_{j+1}\cup \cdots \cup A_{j+s-1}$ of length $s$ $(1\le j\le i-3,\ j+s-1=i)$, that is, a subsequence of annuli between $A_j$ and $A_i$ such that $D_i\supset D_{j-1}$ if $2\le j\le i-3$ or $D_i\supset D_n$ if $j=1$.
Since there exists no turning pair of adjacent annuli, we have $A_j'\supset A_{i+1}',\ A_{j+1}'\supset A_{i+2}',\ \ldots,\ A_{j+s-1}'\supset A_{i+s}'$.
Thus $A_i'\supset A_{i+s}'$ and hence $A_{i+1}\cup\cdots\cup A_{i+s}$ is also a long Reeb of length $s$.
Consequently, we obtain an infinite sequence $A_i'\supset\cdots\supset A_{i+s}'\supset\cdots\supset A_{i+2s}'\supset \cdots $, a contradiction.
\end{proof}

In the conclusion (1) of Lemma \ref{two cases}, there exists a core loop $L$ of the solid torus $W_i$ $(i=1$ or $2)$ such that $L$ intersects $D_i$ in one point for $i=1,\ldots,n$.
Then we have an $n/2$-bridge decomposition of a trivial knot $L$ by $S$ $(n\ge4)$ and it is stabilized by Theorem \ref{Otal}.
This implies that an $n$-bridge decomposition $(B^+,T^+)\cup_{S}(B^-,T^-)$ of $K$ is also stabilized.

In the conclusion (2) of Lemma \ref{two cases}, without loss of generality, we may assume that $(A_1,A_2)$ is a turning pair of adjacent annuli.
Let $E_1\subset B^+$ and $E_2\subset B^-$ be $\partial$-compressing disks for $A_1$ and $A_2$ respectively.

\begin{lemma}\label{essential arcs}
The $\partial$-compressing disk $E_j$ $(j=1,2)$ can be isotoped so that it intersects $A_i$ in essential arcs $(i=3,\ldots,n)$.
\end{lemma}

\begin{proof}
Without loss of generality, we show this lemma only for $E_1$.

We remark first that each annulus $A_i$ $(i=3,5,\ldots,n-1)$ separates the 3-ball $B^+$ into a 3-ball, say $B_i$, and a solid torus, say $V_i$, and $A_i$ is $\partial$-parallel to $A_i'$ in $V_i$ since $A_i$ is a ``1-bridge annulus'', that is, it contains only one essential saddle point and only one maximal point.
We remark next that there are the following three types for an annulus $A_i$ in $V_1$.
\begin{enumerate}
\item $\partial A_i$ consists of mutually non-parallel inessential loops in $A_1'$.
\item $\partial A_i$ consists of mutually parallel inessential loops in $A_1'$.
\item $\partial A_i$ consists of mutually parallel essential loops in $A_1'$.
\end{enumerate}
In this three types, an annulus of type (1) is not $\partial$-parallel to $A_i'$ in $V_1$, and an annulus of type (2) or (3) is $\partial$-parallel to $A_i'$ in $V_1$.
Let $\mathcal{A}$ be the set of annuli contained in $V_1$ from $A_i$ $(i=3,5,\ldots,n-1)$, and $\partial \mathcal{A}\subset \mathcal{A}$ be the set of annuli which are $\partial$-parallel to $A_1'$, in other words, an annulus $A_i$ which satisfies $A_i'\subset A_1'$.

We show this lemma by an induction on $l=|\partial \mathcal{A}|$.

First if $l=0$, then all annuli in $V_1$ are of type (1).
This shows that there exists a string $t_i$ in $B_i$ for all $i$ such that $N(t_i)=B_i$.
Since $A_i$ is a 1-bridge annulus, we can take $t_i$ so that it is also 1-bridge with respect to $h:B^+\to [0,+\infty)$.
Hence $t_i$ bounds a disk $e_i$ with an arc $t_i'$ in $S$ with $\partial t_i=\partial t_i'$ such that $e_i\cap e_j=\emptyset$ for $i\ne j$.
This shows that there exists a $\partial$-compressing disk $E_i$ for $A_i\in \mathcal{A}$ such that $E_i\cap E_j=\emptyset$ for $i\ne j$.
We have particularly $E_1\cap A_i=\emptyset$ for $i=3,5,\ldots,n-1$.

Next suppose that the induction holds for $l=k-1$, and let $l=k$.
Without loss of generality, we may assume that $A_k$ is ``outermost'' in $V_1$, that is, $V_k$ does not contain an annulus in $\partial \mathcal{A}$.
Let $\mathcal{A}_k$ be the set of annuli $A_i\in\mathcal{A}$ which are contained in $B_k$ (including $A_k$), and $\mathcal{A}_k'$ be the set of annuli $A_i\in\mathcal{A}$ which are contained in $V_k$.
Thus $\mathcal{A}=\mathcal{A}_k\cup \mathcal{A}_k'$.
Since $A_k$ is $\partial$-parallel in $V_1$, the height function $h:B^+\to [0,+\infty)$ can be extended to $h':B_k\to [0,+\infty)$, and each annulus $A_i\in \mathcal{A}_k$ is still a 1-bridge annulus with respect to $h'$.
Therefore, by the supposition of the induction for $l=k-1$, there exists a $\partial$-compressing disk $E_1\subset B_k$ for $A_1$ such that $E_1$ intersects each annulus $A_i\in \mathcal{A}_k$ in essential arcs.
Furthermore, $E_1$ can be isotoped so that $\partial E_1$ intersects $A_k\subset \partial B_k$ in essential arcs since $\partial A_k$ consists of mutually parallel loops in $A_1'$ and hence in $\partial B_k\cap V_1$.
On the other hand, since $\mathcal{A}_k'$ does not contain an annulus which is $\partial$-parallel to $A_k'$ in $V_k$, there exists a $\partial$-compressing disk $E_k\subset V_k$ for $A_k$ which is disjoint from $A_i\in \mathcal{A}_k'$ as in the case that $l=0$.
Then, we obtain the desired $\partial$-compressing disk for $A_1$ by gluing $E_1$ and $|\partial E_1\cap A_k|$-copies of $E_k$ along the essential arcs of $\partial E_1\cap A_k$.
\end{proof}

By Lemma \ref{essential arcs}, we have two $\partial$-compressing disks $E_1$ and $E_2$ for $A_1$ and $A_2$ respectively such that $E_j\cap A_i$ consists of essential arcs for $j=1,2$ and $i=3,\ldots n$, and these two disks can be isotoped so that their interiors do not intersect $T^{\pm}$ since $T^{\pm}$ consists of mutually parallel essential arcs or inessential arcs in $A_i$.
This shows that an $n$-bridge decomposition $(B^+,T^+)\cup_{S}(B^-,T^-)$ of $K$ is stabilized.

\subsection{The case that $|F\cap S|=2$}

Next we consider the case that $|F\cap S|=2$.
In this case, $F\cap B^+$ consists of a single annulus $A_1$ and $F\cap B^-$ consists of a single annulus $A_2$ both of which are 1-bridge with respect to $h$.
Therefore the annulus $A_1$ separates $B^+$ into a 3-ball $B_1$ and a solid torus $V_1$, and $A_2$ separates $B^-$ into a 3-ball $B_2$ and a solid torus $V_2$.
It follows that $A_i$ is parallel to $A_i'$ in $V_i$ $(i=1,2)$.
We have $\partial A_1=\partial A_2=C_1\cup C_2$, and $C_1$ and $C_2$ bound disks $D_1$ and $D_2$ in $S$ with $D_1\cap D_2=\emptyset$.

If there exists a string $t_i$ of $T^+$ which is inessential in $A_1$ without loss of generality, then an $n$-bridge decomposition $(B^+,T^+)\cup_{S}(B^-,T^-)$ of $K$ is stabilized.
Indeed, $t_i$ and an arc $t_i'$ in $D_i$ with $\partial t_i=\partial t_i'$ cobound a disk $e_i$ in $B_1$, and an adjacent string $t_{i+1}$ of $T^-$ and an arc $t_{i+1}'$ in $A_2'$ with $\partial t_{i+1}=\partial t_{i+1}'$ cobound a disk $e_{i+1}$ in $V_2$ since $A_2$ is parallel to $A_2'$ in $V_2$.

Otherwise, $T^+$ consists of essential and hence mutually parallel arcs in $A_1$ and $T^-$ also consists of essential and hence mutually parallel arcs in $A_2$.
Since an $n$-bridge position $(B^+,T^+)\cup_{S}(B^-,T^-)$ of $K$ is non-minimal, $T^{\pm}$ consists of $q$-arcs.
In this case, an $n$-bridge decomposition $(B^+,T^+)\cup_{S}(B^-,T^-)$ of $K$ is stabilized as follows.

If we project $T^{\pm}$ into $A_1'=A_2'$ so that it has minimal crossings up to isotopy of $T^{\pm}$ in $A_1'=A_2'$, then we have a $q$-bridge presentation of a $(p,q)$-torus knot.
Then there exists an isotopy of $T^+$ in $B^+$ such that $(q-p)$-pairs of adjacent over/under bridges $(t_i,t_{i+1})$ have no crossing.
Hence we obtain $(q-p)$-destabilizations for $(B^+,T^+)\cup_{S}(B^-,T^-)$.
Furthermore, there exists an isotopy of $T^-$ in $B^-$ which gives a $p$-bridge presentation of a $(q,p)$-torus knot.
For an example on $p=3$ and $q=5$, see Figure \ref{torus knot}.

\begin{figure}[htbp]
	\begin{center}
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.5\linewidth]{torus1.eps}
	\end{center}
	\begin{center}
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.5\linewidth]{torus2.eps}
	\end{center}
	\begin{center}
	\includegraphics[trim=0mm 0mm 0mm 0mm, width=.5\linewidth]{torus3.eps}
	\end{center}
	\caption{A deformation from a $(3,5)$-torus knot to a $(5,3)$-torus knot}
	\label{torus knot}
\end{figure}

This completes the proof of Theorem \ref{torus} and shows that minimal bridge decompositions of a torus knot are unique.
\end{proof}

\bigskip

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