text:
\begin{document}

\title[Positive Mass with Angular Momentum and Charge] {The Positive Mass Theorem with Angular Momentum and Charge for Manifolds with Boundary}

\author[Bryden]{Edward T. Bryden}
\author[Khuri]{Marcus A. Khuri}
\author[Sokolowsky]{Benjamin D. Sokolowsky}
\address{Department of Mathematics\\
Stony Brook University\\
Stony Brook, NY 11794, USA}
\email{ebryden@math.sunysb.edu, khuri@math.sunysb.edu, bsokolowsky@math.sunysb.edu}

\thanks{M. Khuri acknowledges the support of NSF Grant DMS-1708798.}

\begin{abstract}
Motivated by the cosmic censorship conjecture in mathematical relativity,
we establish the precise mass lower bound for an asymptotically flat Riemannian 3-manifold with nonnegative scalar curvature and minimal surface boundary, in terms of angular momentum and charge. In particular this result does not require the restrictive assumptions of simple connectivity and completeness, which are undesirable from both a mathematical and physical perspective.
\end{abstract}
\maketitle

\section{Introduction}
\label{sec1} \setcounter{equation}{0}
\setcounter{section}{1}

Heuristic arguments of Penrose \cite{Penrose} which are well-known to motivate the Penrose inequality \cite{Bray,HuiskenIlmanen}, may also be used to obtain a conjectured lower bound for the ADM mass $m$ of a spacetime in terms of total angular momentum $\mathcal{J}$ and charge $Q$, namely
\begin{equation}\label{0}
m^2 \geq \frac{Q^{2}+  \sqrt{Q^4 + 4\mathcal{J}^2}}{2},
\end{equation}
with equality occurring only for the extreme Kerr-Newman black hole. The arguments leading to \eqref{0} require conservation of angular momentum and charge, and for this it is typically assumed that the spacetime is axisymmetric and satisfies certain conditions related to its matter fields.
Inequality \eqref{0} acts as a necessary condition for the grand cosmological censorship conjecture \cite{Penrose1} as well as the final state conjecture \cite{Klainerman}. Therefore while a counterexample would be detrimental for at least one of these conjectures, confirmation under the most general of possible settings only adds to the prevailing belief in their validity. Moreover, inequality \eqref{0} may be viewed as a refinement of the positive mass theorem \cite{SchoenYau0,Witten} in which a precise contribution to the total mass is given in terms of the rotation and charge of black holes.

The appropriate mathematical setting in which to study this inequality is that of an initial data set $(M,g,k)$, consisting of a Riemannian 3-manifold with metric $g$ and a symmetric 2-tensor $k$ representing the second fundamental form of the embedding into spacetime. An asymptotically flat end is required for the definition of ADM mass, and nonnegative scalar curvature is assumed in order that the positive mass theorem is valid. This curvature condition arises from the physical consideration of nonnegative matter energy density together with a maximal slice $\tr k=0$. Through the combined work of several authors \cite{ch1,ChruscielCosta,ChruscielLiWeinstein,Costa,Dain0,KhuriWeinstein,SchoenZhou}, inequality \eqref{0} has been established when the manifold $(M,g)$ is simply connected, complete, and contains another end which is either asymptotically flat or asymptotically cylindrical. The proof follows a two step procedure, the first of which is to obtain an initial lower bound for the mass in terms a renormalized harmonic map energy. The second consists of minimizing this energy, and showing that the unique global minimizer is the singular harmonic map associated with extreme Kerr-Newman data. Simple connectivity is used in the first step to introduce a specialized coordinate system (Brill coordinates) that allows for a simple bulk integral expression of the mass. Completeness and the asymptotics of the other end also play an important role here in that they prevent the appearance of boundary terms in the formula for the mass. Furthermore, simple connectivity is also used in the second step to ensure the existence of a twist potential to efficiently encode angular momentum and construct the harmonic map energy. Thus, these hypotheses play fundamental roles in the proof and whether they can be removed has been unclear. On the other hand, the Penrose arguments motivating the inequality require no such hypotheses, and for this reason it has been conjectured that these assumptions are unnecessary. They are also unnatural as the positive mass theorem itself does not require such restrictions. In particular, it is important to allow the initial data to have minimal surface boundary, as these may be interpreted as cross-sections of the event horizon. Furthermore, significant generalizations of the positive mass theorem, including the positive mass theorem with charge \cite{GHHP} and the Penrose inequality (with charge) \cite{KhuriWeinsteinYamada}, require a minimal surface boundary to be meaningful. Additionally, from a physical perspective it is not desirable to necessitate the presence of a secondary asymptotic end, as this typically represents the interior of a black hole. Indeed, from the point of view of an outside observer, it is not possible to know the structure of spacetime contained within the event horizon. As for simple connectivity, although topological censorship \cite{FriedmanSchleichWitt} implies that this is an appropriate assumption for initial data within the domain of outer communication, it says nothing about the fundamental group of the interior black hole region. In fact, it suggests that all nontrivial topology is contained within the black hole, and therefore the combined assumptions of simple connectivity, completeness, and the existence of a secondary asymptotically flat end are not physically justified.

The purpose of the present work is to establish \eqref{0} in generality without the unwanted hypotheses discussed above, for a single black hole. We also obtain a mass lower bound in the multi-black hole case consistent with that proved under the more restrictive hypotheses in \cite{ChruscielLiWeinstein,KhuriWeinstein}.

Let vector fields $E$ and $B$ defined on $M$ represent the electric and magnetic fields. It will be assumed that both are divergence free, that is, there is no charged matter. Traces of the Gauss and Codazzi equations for the embedding of the initial data into spacetime yield formulas for the energy and momentum densities of the matter fields minus electromagnetic contributions via the constraint equations
\begin{equation}\label{1}
16\pi\mu_{em} = R+(\tr k)^{2}-|k|^{2}-2(|E|^{2}+|B|^{2}),
\end{equation}
\begin{equation}
8\pi J_{em} = \operatorname{div}(k-(\tr k)g)+2E\times B,

\end{equation}
where $R$ is scalar curvature and $(E\times B)_{i}=\epsilon_{ijl}E^{j}B^{l}$ denotes the cross product in which $\epsilon$ is the volume form of $g$. We will say that the initial data are axially symmetric if there exists a $U(1)$ subgroup within the group of isometries of the Riemannian manifold $(M,g)$, and all relevant quantities are invariant
under the $U(1)$ action. In particular, the following Lie derivatives vanish
\begin{equation}\label{2}
\mathfrak{L}_{\eta}g=\mathfrak{L}_{\eta}k=\mathfrak{L}_{\eta}E=\mathfrak{L}_{\eta}B=0,
\end{equation}
where $\eta$ is the Killing field corresponding to the symmetry.

The initial data will be referred to as asymptotically flat if there exists an end $M_{\text{end}}\subset M$ diffeomorphic to $\mathbb{R}^{3}\setminus\text{Ball}$, such that with respect to the asymptotic coordinates
\begin{equation}\label{falloff1}
g_{ij}=\delta_{ij}+o_{\ell}(r^{-\frac{1}{2}}),\text{ }\text{ }\text{ }\text{ }\partial g_{ij}\in
L^{2}(M_{\text{end}}),\text{
}\text{ }\text{ }
\text{ }k_{ij}=O_{\ell-1}(r^{-\lambda}),\text{ }\text{ }\text{ }\text{ }\mu_{em}, J_{em}^{i},J_{em}(\eta)\in L^{1}(M_{\text{end}}),
\end{equation}
\begin{equation}\label{falloff2}
E^{i}=O_{\ell-1}(r^{-\lambda}),\text{ }\text{ }\text{ }\text{ }\text{ }B^{i}=O_{\ell-1}(r^{-\lambda}),\text{
}\text{ }\text{ }
\text{ }\lambda>\frac{3}{2},
\end{equation}
for some $\ell\geq 5$.
These fall-off conditions ensure that the ADM mass and angular momentum, as well as the total electric and magnetic charge are well-defined by
\begin{equation}\label{admmass}
m=\frac{1}{16\pi}\int_{S_{\infty}}(g_{ij,i}-g_{ii,j})\nu^{j},
\end{equation}
\begin{equation}\label{admam}
\mathcal{J}=\frac{1}{8\pi}\int_{S_{\infty}}(k_{ij}-(\tr k)g_{ij})\nu^{i}\eta^{j},
\end{equation}
\begin{equation}\label{charge}
Q_{e} = \frac1{4\pi} \int_{S_\infty} E_i \nu^i\, , \qquad
Q_{b} = \frac1{4\pi} \int_{S_\infty} B_i \nu^i\, .
\end{equation}
Here limits as $r\rightarrow\infty$ for integrals taken over coordinate spheres $S_r$, with unit outer normal $\nu$, are represented by $S_{\infty}$. The total charge is the combination of the total electric and magnetic charges $Q^2=Q_{e}^2+Q_{b}^2$. It should also be pointed out that \eqref{admmass} is referred to as mass by an abuse of terminology, as this expression is technically the ADM energy or time component in the ADM 4-momentum. Furthermore an electromagnetic contribution to the definition of angular momentum is sometimes included \cite{DainKhuriWeinsteinYamada}, although the two definition agree under the current asymptotics.

In order for the Penrose heuristic arguments to be valid an energy condition is needed so that Hawking's area theorem \cite{Hawking} may be utilized, and hypotheses must be imposed which guarantee the conservation of angular momentum and charge. This motivates the following assumptions used in the proof of \eqref{0} from the initial data perspective. For the energy condition, nonnegative nonelectromagnetic matter energy density $\mu_{em}\geq 0$ will be assumed. This is consistent with the dominant energy condition typically used in association with the positive mass theorem (including charge), as well as previous inequalities involving angular momentum. Furthermore, axisymmetry and the electrovacuum assumption $\mu_{em}=|J_{em}|=0$ have been used in the past \cite{Dain} with regards to conservation of angular momentum and charge. However, here it is only necessary to require vanishing of one component of the momentum density, namely in the Killing direction $J_{em}(\eta)=0$, and the divergence free property for the Maxwell field. Inequality \eqref{0} is known to be false without the assumption of axisymmetry \cite{HuangSchoenWang}.

\begin{theorem}\label{massam}
Let $(M,g,k,E,B)$ be an axisymmetric, maximal initial data set for the Einstein-Maxwell equations with one asymptotically flat end, minimal surface boundary, and satisfying $\mu_{em}\geq 0$ in addition to $J_{em}(\eta)=\operatorname{div}E=\operatorname{div}B=0$. If either
\begin{itemize}
\item [(i)] the outermost minimal surface has a single component, or\smallskip

\item [(ii)] the boundary $\partial M$ has one component and $M$ is simply connected,
\end{itemize}
then
\begin{equation}\label{3}
m^2>\frac{Q^{2}+\sqrt{Q^{4}+4\mathcal{J}^{2}}}{2}.
\end{equation}
\end{theorem}

The first point to note is that there is a strict inequality in \eqref{3}. This is to be expected since from the heuristic physical arguments leading to \eqref{0}, equality should only be achieved if the initial data agree with the canonical slice of an extreme Kerr-Newman spacetime. However the extreme Kerr-Newman data do not possess a minimal surface, but rather have a cylindrical end, and therefore do not satisfy the hypotheses of Theorem \ref{massam}. The minimal surface boundary, which could consist of many components, together with the asymptotically flat end guarantee the existence of an outermost minimal surface \cite{Eichmair}, and the assumption that it has one component is analogous to the case of the Penrose inequality treated by Huisken and Ilmanen \cite{HuiskenIlmanen}. In order to treat \eqref{3} in the presence of a multicomponent outermost minimal surface it is most likely that new ideas will be needed, as was the case for the multiple black hole version of the Penrose inequality established by Bray \cite{Bray}. It is interesting to note that unlike the Penrose inequality, \eqref{3} continues to hold if the boundary $\partial M$ is merely a single component minimal surface but not necessarily outer minimizing. In fact under this assumption treated in $(ii)$, one can drop the hypothesis on the outermost minimal surface and replace it with simple connectivity of $M$ to obtain the same conclusion.

The proof of Theorem \ref{massam} is based on a doubling procedure in so called pseudospherical coordinates, where the data are reflected across the outermost minimal surface. This requires axisymmetry of the outermost minimal surface, a fact that does not appear to exist within the literature and is thus proven below in Proposition \ref{axi}. This result is of independent interest as it may be applied elsewhere, for example to extensions of the Penrose inequality that include contributions from angular momentum.

We are able to extend Theorem \ref{massam} to allow for certain types of multiple black holes by including a mixture of boundary components and extra asymptotically flat as well as asymptotically cylindrical ends (see \cite{KhuriWeinstein}). However, as in the case of a complete, simply connected initial data set, the presence of multiple black holes does not immediately yield an explicit expression for the mass lower bound \cite{ChruscielLiWeinstein,KhuriWeinstein}. Rather, the lower bound is given in terms of the reduced harmonic energy of a Weinstein stationary solution \cite{weinstein96} to the Einstein-Maxwell equations having the same angular momentum and charge for each black hole. This harmonic energy is denoted by $\mathcal{F}$, and is a function of the angular momenta and charge. It is conjectured that the resulting inequality coincides with the expression \eqref{3} in which $\mathcal{J}$ and $Q$ are the sums of the angular momenta and charge from the different horizon components.

\begin{theorem}\label{thm2}
Let $(M,g,k,E,B)$ be an axisymmetric, maximal, asymptotically flat initial data set for the Einstein-Maxwell equations having a minimal surface boundary and a finite number of additional ends each of which is asymptotically flat or asymptotically cylindrical. Assume further that $\mu_{em}\geq 0$ in addition to $J_{em}(\eta)=\operatorname{div}E=\operatorname{div}B=0$.
If either
\begin{itemize}
\item [(i)] at most one component of the outermost minimal surface encloses components of the boundary $\partial M$ or nonsimply connected domains, or\smallskip

\item [(ii)] the boundary $\partial M$ has one component and $M$ is simply connected,
\end{itemize}
then
\begin{equation}\label{3.1}
m\geq\mathcal{F}(\mathcal{J}_{1},\ldots,\mathcal{J}_{N},Q_{e}^{1},\ldots,Q_{e}^{N},
Q_{b}^{1},\ldots,Q_{b}^{N})
\end{equation}
where $N$ is the combined number of additional ends and components of $\partial M$ and $\mathcal{J}_i$, $Q_{e}^i$, $Q_{b}^i$ represent the angular momentum and charge associated with each of these ends and boundary components.
\end{theorem}

\section{The Doubling Procedure in Pseudospherical Coordinates}
\label{sec2} \setcounter{equation}{0}
\setcounter{section}{2}

Consider the setting of case $(ii)$ in Theorem \ref{massam} where $(M,g)$ is axisymmetric, asymptotically flat, and simply connected with a single component minimal surface boundary. It follows from \cite[Theorem 2.2]{ChruscielNguyen} that $M$ is diffeomorphic to $\mathbb{R}^3\setminus B_{m_{1}/2}(0)$, and there exists a global system of cylindrical-type coordinates $(\rho,z,\phi)$ on this domain such that the metric takes the form
\begin{equation}\label{metric}
g=e^{-2U+2\alpha}(d\rho^2+dz^2)+\rho^2 e^{-2U}(d\phi+A_{\rho} d\rho +A_{z} dz)^2,
\end{equation}
where the Killing field $\eta=\partial_{\phi}$. The isothermal part of \eqref{metric} is the metric on the orbit space $M/U(1)$, and the remaining part arises from the dual 1-form $\rho^2 e^{-2U}(d\phi +A_{\rho} d\rho +A_{z} dz)$ to the Killing field. This structure for $g$ in pseudospherical coordinates is related to the Weyl-Papapetrou form \cite{ChruscielNguyen} used in the reduction of the stationary axisymmetric vacuum Einstein equations. With the standard transformation $\rho=r\sin\theta$, $z=r\cos\theta$ producing spherical-type coordinates $(r,\theta,\phi)$ with ranges $m_{1}/2\leq r<\infty$, $0\leq\theta\leq \pi$, and $0\leq\phi<2\pi$, the fall-off of the metric coefficients in the asymptotically flat end is given by
\begin{equation}\label{f}
U=o_{\ell-3}(r^{-\frac{1}{2}}),\quad \alpha=o_{\ell-4}(r^{-\frac{1}{2}}),\quad
A_{\rho}=\rho o_{\ell-3}(r^{-\frac{5}{2}}),\quad A_{z}=o_{\ell-3}(r^{-\frac{3}{2}}).
\end{equation}
Furthermore $\alpha=0$ on the axis $\rho=0$, and all coefficients are independent of $\phi$. Note also that the value $m_1>0$ is uniquely determined, and the existence of pseudospherical coordinates does not require the boundary $\partial M$ to be minimal. Since the mean curvature of a coordinate sphere $S_r$ is
\begin{equation}
H=\frac{2/r+\partial_{r}(\alpha-2U)}{\sqrt{e^{-2U+2\alpha}+\rho^2 e^{-2U} A_{r}^2}}
\end{equation}
where $A_{r}=\sin\theta A_{\rho}+\cos\theta A_{z}$, the assumption of a minimal boundary $\partial M=S_{m_{1}/2}$ is equivalent to
\begin{equation}\label{minimal}
\partial_{r}\left(U-\frac{1}{2}\alpha\right)=\frac{2}{m_1}.
\end{equation}

A particularly advantageous feature of the metric structure \eqref{metric} is the simple expression obtained for the scalar curvature \cite{Dain0}
\begin{equation}\label{scalar}
2e^{-2U+2\alpha}R=8\Delta U-4\Delta_{\rho,z}\alpha-4|\nabla U|^{2}-\rho^{2}e^{-2\alpha}
\left(A_{\rho,z}-A_{z,\rho}\right)^{2},
\end{equation}
where $\Delta$ is the Laplacian on $\mathbb{R}^{3}$ with respect to the flat metric $\delta=d\rho^2+dz^2+\rho^2 d\phi^2$ and $\Delta_{\rho,z}=\partial_{\rho}^{2}+\partial_{z}^{2}$. Moreover the constraint equation \eqref{1}, and the assumptions of a maximal slice $\tr k=0$ and nonnegative energy density $\mu_{em}\geq 0$ imply that
\begin{align}\label{scalar1}
\begin{split}
R=&16\pi\mu_{em}+|k|^{2}+2\left(|E|^{2}+|B|^{2}\right)\\
\geq
&2\frac{e^{6U-2\alpha}}{\rho^{4}}|\nabla v
+\chi\nabla\psi-\psi\nabla\chi|^{2}
+2\frac{e^{4U-2\alpha}}{\rho^{2}}\left(|\nabla\chi|^{2}
+|\nabla\psi|^{2}\right),
\end{split}
\end{align}
where $v$, $\chi$, and $\psi$ are potential functions for angular momentum, electric charge, and magnetic charge respectively. More precisely, the divergence free property of the electric and magnetic fields combined with Cartan's magic formula shows that the 1-forms $\iota_{\eta}\star E$ and $\iota_{\eta}\star B$ are closed, where $\iota$ and $\star$ denote interior product and the Hodge star operation. Hence simple connectivity yields global potentials satisfying
\begin{equation}\label{empotential}
d\chi=\iota_{\eta}\star E,\quad\quad\quad d\psi=\iota_{\eta}\star B.
\end{equation}
Furthermore, as shown in \cite{KhuriWeinstein} the 1-form $2\star( k(\eta)\wedge \eta)-\chi d\psi+\psi d\chi$ is closed exactly when $J_{em}(\eta)=0$. Therefore under the hypotheses of Theorem \ref{massam} there exists a global twist potential satisfying
\begin{equation}\label{twistpotential}
dv=2\star( k(\eta)\wedge \eta)-\chi d\psi+\psi d\chi.
\end{equation}
The inequality in \eqref{scalar1} then follows in a straightforward way from \eqref{empotential} and \eqref{twistpotential}.  Moreover, if
$\omega=dv+\chi d\psi-\psi d\chi$ then asymptotics \cite{KhuriWeinstein} for the potentials are expressed by
\begin{equation}\label{asym}
|\omega|=\rho^{2}O(r^{-\lambda}),\text{ }\text{ }\text{ }\text{ }|\nabla\chi|+|\nabla\psi|=\rho O(r^{-\lambda})\text{
}\text{ }\text{ as }\text{ }\text{ }r\rightarrow\infty,
\end{equation}
\begin{equation}\label{asym.1}
|\omega|=O(\rho^{2}),\text{ }\text{ }\text{ }\text{ }|\nabla\chi|+|\nabla\psi|=O(\rho) \text{ }\text{ }\text{ as }\text{
}\text{ }\rho\rightarrow 0\text{ }\text{ }\text{ in }\text{ }\text{ }\mathbb{R}^3\setminus B_{m_1/2}(0).
\end{equation}
In addition, since $|\eta|=0$ on the $z$-axis all the potential functions are constant there, and the difference of these constants associated with the two connected components $I_{+}=\{\rho=0, z> m_{1}/2\}$ and $I_{-}=\{\rho=0, z<-m_{1}/2\}$ of the axis yield the angular momentum and charges
\begin{equation}\label{potentialcharge}
\mathcal{J}=\frac{1}{4}\left(v|_{I_{-}}-v|_{I_{+}}\right),\quad\quad
Q_e=\frac{1}{2}\left(\chi|_{I_{-}}-\chi|_{I_{+}}\right),\quad\quad
Q_b=\frac{1}{2}\left(\psi|_{I_{-}}-\psi|_{I_{+}}\right).
\end{equation}

Typically a mass lower bound in terms of a harmonic map energy is obtained by
integrating \eqref{scalar} over $M$ and applying a version of \eqref{scalar1}. This works well when $(M,g)$ is complete, however here the presence of a boundary leads to boundary terms which are not desirable when minimizing the harmonic map energy. Therefore we seek to double the manifold across its boundary, and show that the same strategy may be carried out on the doubled manifold with two ends. The primary difficulty arises from the lack of regularity across the doubling surface. Nevertheless we show that the minimal surface hypothesis is sufficient for the argument to go through.

Consider the conformal map $f:B_{m_1/2}\setminus\{0\}\rightarrow \mathbb{R}^3\setminus B_{m_1/2}$ given by spherical inversion
\begin{equation}
f(\tilde{r},\tilde{\theta},\tilde{\phi})=\left(\left(\frac{m_1}{2}\right)^2\frac{1}{\tilde{r}},
\tilde{\theta},\tilde{\phi}\right),
\end{equation}
which is expressed in cylindrical coordinates as
\begin{equation}\label{b}
\rho=\left(\frac{m_1}{2}\right)^2\frac{\tilde{\rho}}{\tilde{r}^2},\quad\quad\quad
z=\left(\frac{m_1}{2}\right)^2\frac{\tilde{z}}{\tilde{r}^2},\quad\quad\quad
\phi=\tilde{\phi}.
\end{equation}
Pulling back the metric to $B_{m_1/2}\setminus\{0\}$ yields
\begin{equation}\label{newmetric}
\tilde{g}:=f^{*}g=e^{-2\tilde{U}+2\tilde{\alpha}}
(d\tilde{\rho}^2+d\tilde{z}^2)+\tilde{\rho}^2 e^{-2\tilde{U}}(d\tilde{\phi}
+\tilde{A}_{\rho} d\tilde{\rho} +\tilde{A}_{z} d\tilde{z})^2,
\end{equation}
where
\begin{equation}\label{a}
\tilde{U}=2\log\tilde{r}+2\log(2/m_1)+U\circ f,\quad\quad\quad\quad
\tilde{\alpha}=\alpha\circ f,
\end{equation}
\begin{equation}
\tilde{A}_{\rho}=\tilde{r}^{-4}\left(\frac{2}{m_1}\right)^2
\left[(\tilde{z}^2-\tilde{\rho}^2)A_{\rho}-2\tilde{\rho}A_{z}\right],\quad\quad
\tilde{A}_{z}=\tilde{r}^{-4}\left(\frac{2}{m_1}\right)^2
\left[(\tilde{\rho}^2-\tilde{z}^2)A_{z}-2\tilde{z}A_{\rho}\right].
\end{equation}
This leads to a metric and potentials globally defined on the complement of the origin
\begin{equation}
\bar{g}=
  \begin{cases}
  g & \text{on $\mathbb{R}^3\setminus B_{m_1/2}$,} \\
  \tilde{g} & \text{on $B_{m_1/2}\setminus\{0\}$.}
  \end{cases}
\end{equation}
Similarly, the potentials may also be extended to the ball by setting $\tilde{v}=v\circ f$, $\tilde{\chi}=\chi\circ f$, $\tilde{\psi}=\psi\circ f$ in $B_{m_1/2}\setminus\{0\}$, and the corresponding functions defined on $\mathbb{R}^3\setminus\{0\}$ will be denoted $\bar{v}$, $\bar{\chi}$, and $\bar{\psi}$. These functions and the metric $\bar{g}$ are $C^{0,1}$ and smooth away from the reflection sphere $S_{m_1/2}$.

The form of the metric \eqref{newmetric} guarantees that the scalar curvature of $\bar{g}$ satisfies the equation \eqref{scalar} on all of $\mathbb{R}^3\setminus\{0\}$. Moreover it also satisfies the lower bound in \eqref{scalar1}. To see this observe that
\begin{equation}
|\nabla\chi|^2\circ f=
(\partial_{\rho}\chi)^2\circ f+(\partial_{z}\chi)^2\circ f
=\left(\frac{2}{m_1}\right)^4\tilde{r}^4\left[(\partial_{\tilde{\rho}}\tilde{\chi})^2
+(\partial_{\tilde{z}}\tilde{\chi})^2\right]
=\left(\frac{2}{m_1}\right)^4\tilde{r}^4|\tilde{\nabla}\tilde{\chi}|^2,
\end{equation}
and similarly
\begin{equation}
|\nabla v+\chi\nabla\psi-\psi\nabla\chi|^2\circ f
=\left(\frac{2}{m_1}\right)^4\tilde{r}^4|\tilde{\nabla} \tilde{v}+\tilde{\chi}\tilde{\nabla}\tilde{\psi}-\tilde{\psi}\tilde{\nabla}
\tilde{\chi}|^2.
\end{equation}
Combining this with \eqref{scalar1}, \eqref{b}, and \eqref{a} shows that
in $B_{m_1/2}\setminus\{0\}$
\begin{align}\label{scalarfinal}
\begin{split}
\tilde{R}=&R\circ f\\
\geq
&2\frac{e^{6U\circ f-2\alpha\circ f}}{(\rho\circ f)^{4}}|\nabla v
+\chi\nabla\psi-\psi\nabla\chi|^{2}\circ f
+2\frac{e^{4U\circ f-2\alpha\circ f}}{(\rho\circ f)^{2}}\left(|\nabla\chi|^{2}\circ f
+|\nabla\psi|^{2}\circ f\right)\\
=& 2\frac{e^{6\tilde{U}-2\tilde{\alpha}}}{\tilde{\rho}^{4}}|\tilde{\nabla} \tilde{v}
+\tilde{\chi}\tilde{\nabla}\tilde{\psi}-\tilde{\psi}\tilde{\nabla}\tilde{\chi}|^{2}
+2\frac{e^{4\tilde{U}-2\tilde{\alpha}}}{\tilde{\rho}^{2}}
\left(|\tilde{\nabla}\tilde{\chi}|^{2}
+|\tilde{\nabla}\tilde{\psi}|^{2}\right).
\end{split}
\end{align}
It follows that the scalar curvature of the doubled metric $\bar{g}$ satisfies the desired lower bound on $\mathbb{R}^3\setminus\{0\}$ away from the sphere $S_{m_1/2}$.
Although the metric is not sufficiently regular across this sphere to have a pointwise defined scalar curvature on this surface, the fact that it is a minimal surface with respect to both inner and outer domains guarantees that $\bar{R}$ satisfies the inequality distributionally. Furthermore the minimal surface property allows for the fundamental mass lower bound in terms of scalar curvature, despite the lack of metric regularity.

In order to establish the mass lower bound it is necessary to note that the doubled manifold $(\bar{M},\bar{g})$, where $\bar{M}=\mathbb{R}^{3}\setminus\{0\}$, possesses two asymptotically flat ends. Indeed, at the additional end near the origin the metric coefficients and potentials satisfy the asymptotics
\begin{equation}\label{asym1}
\bar{U}=2\log r+C+o_{1}(r^{\frac{1}{2}}),\text{ }\text{ }\text{ }\text{ }\bar{\alpha}=o_{1}(r^{\frac{1}{2}}),\text{
}\text{ }\text{ }\text{ }\bar{A}_{\rho}=\rho o_{1}(r^{-\frac{5}{2}}),\text{ }\text{ }\text{ }\text{
}\bar{A}_{z}=o_{1}(r^{-\frac{3}{2}}),
\end{equation}
\begin{equation}\label{asym2}
|\bar{\omega}|=\rho^{2}O(r^{\lambda-6}),\text{ }\text{ }\text{ }|\nabla\bar{\chi}|+|\nabla\bar{\psi}|=\rho O(r^{\lambda-4})\text{
}\text{ }\text{ as }\text{ }\text{ }r\rightarrow 0,
\end{equation}
for some constant $C$. Here and in what follows, unless stated otherwise, the tilde notation will be removed from coordinates within the domain $B_{m_1/2}\setminus\{0\}$.

\begin{lemma}\label{lemma1}
The doubled manifold $(\bar{M},\bar{g})$ possesses two asymptotically flat ends, and the mass is given by
\begin{equation}\label{masslower}
m
=\frac{1}{32\pi}\int_{\mathbb{R}^{3}}\left(2e^{-2\bar{U}+2\bar{\alpha}}\!\text{ }\bar{R}
+4|\nabla \bar{U}|^{2}
+\rho^{2}e^{-2\bar{\alpha}}
(\bar{A}_{\rho,z}-\bar{A}_{z,\rho})^{2}
\right)dx,
\end{equation}
where $dx$ is the Euclidean volume element.
\end{lemma}

\begin{proof}
Although the metric $\bar{g}$ is only Lipschitz across $S_{m_1/2}$, the fact that this sphere is a minimal surface guarantees that a particular combination of coefficients has improved regularity, namely $\bar{U}-\frac{1}{2}\bar{\alpha}\in C^{1,1}$. Moreover, this is all that is needed to establish \eqref{masslower}.

First observe that in light of \eqref{minimal}
\begin{equation}\label{agree1}
\lim_{r\rightarrow \frac{m_1}{2}^+} \partial_{r}\left(\bar{U}-\frac{1}{2}\bar{\alpha}\right)=\frac{2}{m_1}.
\end{equation}
It suffices then to show that the limit from inside $B_{m_1/2}$ yields the same value. For emphasis we will use the tilde notation to perform this computation. By
\eqref{a}
\begin{equation}
\partial_{\tilde{r}}\tilde{U}=\frac{2}{\tilde{r}}+\partial_{r}U\frac{\partial r}{\partial\tilde{r}}
=\frac{2}{\tilde{r}}-\left(\frac{m_1}{2}\right)^2 \frac{1}{\tilde{r}^2}\partial_{r}U.
\end{equation}
Therefore \eqref{minimal} implies
\begin{equation}
\partial_{\tilde{r}}\tilde{U}|_{\tilde{r}=\frac{m_1}{2}}=
\frac{4}{m_1}-\partial_{r}U|_{r=\frac{m_1}{2}}
=\frac{2}{m_1}-\frac{1}{2}\partial_{r}\alpha|_{r=\frac{m_1}{2}}.
\end{equation}
On the other hand
\begin{equation}
\partial_{\tilde{r}}\tilde{\alpha}=-\left(\frac{m_1}{2}\right)^2 \frac{1}{\tilde{r}^2}\partial_{r}\alpha,
\end{equation}
and therefore the desired conclusion follows
\begin{equation}\label{agree}
\lim_{r\rightarrow \frac{m_1}{2}^-} \partial_{r}\left(\bar{U}-\frac{1}{2}\bar{\alpha}\right)=\frac{2}{m_1}.
\end{equation}

We will now show that \eqref{masslower} holds. According to \cite{ch1}
\begin{equation}\label{fk}
m=\lim_{r\to \infty}\frac{1}{4\pi}\left(\int_{S_r}\partial_r\left(\bar{U}
-\frac{1}{2}\bar{\alpha}\right)+\frac{1}{2}\int_{W_r}\frac{\bar{\alpha}}{\rho} \right),
\end{equation}
where $W_r=\{\rho=r, -r<z<r\}$ is the wall of the cylinder or radius $r$.
Next observe that \eqref{agree1} and \eqref{agree} yield
\begin{align}\label{ibp}
\begin{split}
\int_{S_r}\partial_r\left(\bar{U}-\frac{1}{2}\bar{\alpha}\right) =&\int_{B_r\setminus B_{m_1/2}}\Delta\left(\bar{U}-\frac{1}{2}\bar{\alpha}\right)dx +\lim_{r\rightarrow \frac{m_1}{2}^+}\int_{S_{r}}\partial_{r}\left( \bar{U}-\frac{1}{2}\bar{\alpha}\right)
\\
=&\int_{B_r\setminus B_{m_1/2}}\Delta\left(\bar{U}-\frac{1}{2}\bar{\alpha}\right)dx +
\lim_{r\rightarrow \frac{m_1}{2}^-}\int_{S_{r}}\partial_{r}\left( \bar{U}-\frac{1}{2}\bar{\alpha}\right)
\\
=&\int_{B_r\setminus B_{m_1/2}}\Delta\left(\bar{U}-\frac{1}{2}\bar{\alpha}\right)dx
+\int_{B_{m_1/2}}\Delta\left(\bar{U}-\frac{1}{2}\bar{\alpha}\right)dx,
\end{split}
\end{align}
since the asymptotics \eqref{asym1} show that
\begin{equation}
\lim_{r\rightarrow 0}\int_{S_{r}}\partial_{r}\left( \bar{U}-\frac{1}{2}\bar{\alpha}\right)=0.
\end{equation}
Moreover, since $\bar{\alpha}$ is continuous across $S_{m_1/2}$, vanishes away from the origin on the $z$-axis, and satisfies \eqref{asym1}
\begin{equation}\label{al}
\lim_{r\rightarrow\infty}\int_{W_r}\frac{\bar{\alpha}}{\rho}
=\int_{\mathbb{R}^3}\frac{1}{\rho}\partial_{\rho}\bar{\alpha}dx.
\end{equation}
Finally, since \eqref{scalar} holds globally on $\bar{M}$ we have
\begin{equation}\label{uu}
\Delta\left(\bar{U}-\frac{1}{2}\bar{\alpha}\right)
+\frac{1}{2\rho}\partial_{\rho}\bar{\alpha}
=\Delta\bar{U}-\frac{1}{2}\Delta_{\rho,z}\bar{\alpha}
=\frac{1}{4}e^{-2\bar{U}+2\bar{\alpha}}\bar{R}
+\frac{1}{2}|\nabla\bar{U}|^2
+\frac{1}{8}\rho^2 e^{-2\bar{\alpha}}
\left(\bar{A}_{\rho,z}-\bar{A}_{z,\rho}\right)^2.
\end{equation}
The desired mass formula \eqref{masslower} now follows by combining
\eqref{fk}, \eqref{ibp}, \eqref{al}, and \eqref{uu}.
\end{proof}

Lemma \ref{lemma1} relates the mass to an energy functional with the help of
\eqref{scalar1} and \eqref{scalarfinal}. Namely together they imply
\begin{equation}\label{190}
m\geq\mathcal{I}(\Psi),
\end{equation}
where $\Psi=(\bar{U},\bar{v},\bar{\chi},\bar{\psi})$ and
\begin{equation}\label{energy}
\mathcal{I}(\Psi)
=\frac{1}{8\pi}\int_{\mathbb{R}^{3}}\left(|\nabla \bar{U}|^{2}
+\frac{e^{4\bar{U}}}{\rho^{4}}|\nabla \bar{v}
+\bar{\chi}\nabla\bar{\psi}-\bar{\psi}\nabla\bar{\chi}|^{2}
+\frac{e^{2\bar{U}}}{\rho^{2}}\left(|\nabla\bar{\chi}|^{2}
+|\nabla\bar{\psi}|^{2}\right)\right)dx.
\end{equation}
The functional $\mathcal{I}$ may be interpreted as the reduced harmonic energy \cite{KhuriWeinstein} for maps $\Psi:\mathbb{R}^3\setminus\{0\}\rightarrow
\mathbb{H}_{\mathbb{C}}^2$ into the complex hyperbolic plane.
Note that the asymptotics \eqref{f}, \eqref{asym}, \eqref{asym1}, and \eqref{asym2} guarantee that $\mathcal{I}(\Psi)$ is finite precisely when $\lambda>\frac{3}{2}$.

\begin{proof}[Proof of Theorem \ref{massam} $(ii)$]
Since the map $\Psi$ is smooth away from the sphere $S_{m_1/2}$, Lipschitz across this surface, and satisfies the asymptotics \eqref{f}, \eqref{asym}, \eqref{asym.1}, \eqref{asym1}, \eqref{asym2}, the gap bound of Schoen and Zhou \cite{SchoenZhou} applies to yield
\begin{equation}\label{gapbound}
\mathcal{I}(\Psi)-\mathcal{I}(\Psi_{0})
\geq C\left(\int_{\mathbb{R}^{3}}
\operatorname{dist}_{\mathbb{H}_{\mathbb{C}}^{2}}^{6}(\Psi,\Psi_{0})dx
\right)^{\frac{1}{3}},
\end{equation}
where $\Psi_0$ is the renormalized harmonic map associated with the extreme Kerr-Newman black hole possessing the same angular momentum and charge as $\Psi$,
and $\operatorname{dist}_{\mathbb{H}_{\mathbb{C}}}$ denotes distance in the complex hyperbolic plane. In particular, together with \eqref{190} we obtain
\begin{equation}
m\geq\mathcal{I}(\Psi_0).
\end{equation}
The desired inequality \eqref{3} now follows since
\begin{equation}\label{kerrnewman}
\mathcal{I}(\Psi_0)^2=\frac{Q^{2}+\sqrt{Q^{4}+4\mathcal{J}^{2}}}{2}.
\end{equation}

Consider now the case in which equality holds in \eqref{3}. This implies, with the help of \eqref{190}, \eqref{gapbound}, and \eqref{kerrnewman}, that $\Psi=\Psi_0$ and in particular $U=U_0$. However this is a contradiction since the asymptotics  \eqref{asym1} show that $U=2\log r+O(1)$ as $r\rightarrow 0$, whereas the corresponding asymptotics for the extreme Kerr-Newman map are given by $U_0=\log r +O(1)$. This difference arises from the fact that $\Psi$ arises from an asymptotically flat geometry near the origin, while the extreme Kerr-Newman initial data possess instead an asymptotically cylindrical end in this location.
\end{proof}

\section{The Outermost Minimal Surface in Axisymmetry}
\label{sec3} \setcounter{equation}{0}
\setcounter{section}{3}

Let $(M,g)$ be as in case $(i)$ in Theorem \ref{massam}. Since this manifold is asymptotically flat and possesses a minimal surface boundary, there exists a unique outermost minimal surface \cite{Eichmair,HuiskenIlmanen} which is a compact embedded smooth hypersurface $\Sigma$. The term outermost refers to the fact that there are no other minimal surfaces homologous to $\Sigma$ which lie outside it with respect to the asymptotic end. The set $M\setminus \Sigma$ consists of one unbounded component, and perhaps several bounded components the union of which will be denoted by $\Omega$, so that $\partial\Omega=\Sigma$.  Each component of the outermost minimal surface must be a topological 2-sphere \cite{Galloway}, since $\Sigma$ is outer area minimizing in that it has the least area among all surfaces which enclose it. Furthermore, according to \cite[Lemma 4.1]{HuiskenIlmanen} $M\setminus \Sigma$ is diffeomorphic to the complement of a finite number of open 3-balls in $\mathbb{R}^3$ with disjoint closure. Here we show that the property of axisymmetry for the ambient manifold descends to $\Sigma$.

\begin{prop}\label{axi}
If $(M,g)$ is axisymmetric then the outermost minimal surface is also axisymmetric.
\end{prop}

\begin{proof}
Suppose that the outermost minimal surface $\Sigma$ is not axisymmetric. Let $\varphi_t$ denote the flow of the axisymmetric Killing field $\eta$, so that $\partial_t\varphi_t=\eta\circ\varphi_t$. Lack of axisymmetry implies that $\eta$ is not tangential to $\Sigma$ at all points. Therefore there exists a nonzero $t_0\sim 0$ such that a domain within $\varphi_{t_0}(\Sigma)$ lies outside of $\Sigma$. Note that $\varphi_t$ is a flow by isometries so $\varphi_{t_0}(\Sigma)$ is a minimal surface, and it is still an embedded 2-sphere.

Let $\mathcal{S}$ denote the compact set which is the union of all compact immersed minimal surfaces within $M$, and define the trapped region $\mathcal{T}$ to be the union of $\mathcal{S}$ with all the bounded components of $M\setminus\mathcal{S}$.
The trapped region is compact and its topological boundary is comprised of embedded smooth minimal 2-spheres \cite[Lemma 4.1]{HuiskenIlmanen}. In fact the outermost minimal surface arises as the boundary $\partial\mathcal{T}$.
In light of this, and the fact that a portion of $\varphi_{t_0}(\Sigma)$ lies outside $\Sigma$, it follows that $\Sigma\neq\partial\mathcal{T}$ contradicting the uniqueness of the outermost minimal surface.
\end{proof}

\begin{proof}[Proof of Theorem \ref{massam} $(i)$]
Assume that the outermost minimal surface $\Sigma$ has a single component. From the discussion above we then have that $M_0 =M\setminus\Omega$ is diffeomorphic to the complement of a 3-ball in $\mathbb{R}^3$. Proposition \ref{axi} guarantees that $\partial M_0$ is axisymmetric, and hence $M_0$ is axisymmetric. It follows that $(M_0,g,k,E,B)$ satisfies the hypotheses of Theorem \ref{massam} $(ii)$, and has the same mass, angular momentum, and charge as the original data.
Part $(ii)$ may now be applied to obtain \eqref{3}.
\end{proof}

\section{Multiple Black Holes}
\label{sec4} \setcounter{equation}{0}
\setcounter{section}{4}

Consider the setting of case $(ii)$ in Theorem \ref{thm2} where $(M,g)$ is axisymmetric, asymptotically flat, simply connected, with a single component minimal surface boundary, and a finite number $n$ of additional ends each of which is asymptotically flat or asymptotically cylindrical; see \cite{KhuriWeinstein} for a definition of asymptotically cylindrical ends. The additional ends may be interpreted physically as individual black holes. A version of pseudospherical coordinates exists for this situation, where each additional end is represented by a puncture on the $z$-axis and again the boundary component is identified with a coordinate sphere.

\begin{prop}
Under the hypotheses of Theorem \ref{thm2} $(ii)$, $M$ is diffeomorphic to $\left(\mathbb{R}^3\setminus B_{m_1/2}(0)\right)\setminus\cup_{i=1}^{n}\{p_i\}$ and there exists a global system of cylindrical-type coordinates $(\rho,z,\phi)$ such that $g$ takes the form \eqref{metric} with $\alpha=0$ whenever $\rho=0$. Each puncture $p_i$ represents an additional end in which the metric coefficients have the asymptotics \eqref{asym1} in the asymptotically flat case, or
\begin{equation}\label{asym3}
U=\log r_i +O_{1}(1),\text{ }\text{ }\text{ }\text{ }\alpha=o_{1}(r_{i}^{\frac{1}{2}}),\text{
}\text{ }\text{ }\text{ }A_{\rho}=\rho o_{1}(r_i^{-\frac{5}{2}}),\text{ }\text{ }\text{ }\text{
}A_{z}=o_{1}(r_i^{-\frac{3}{2}}),
\end{equation}
in the asymptotically cylindrical case. Here $r_i$ denotes the Euclidean distance to the puncture $p_i$.
\end{prop}

\begin{proof}
The proof is nearly identical to that of \cite[Theorem 2.2]{ChruscielNguyen}, and thus we only give an outline. Since $M$ is simply connected the single boundary component $\partial M$ must topological be a 2-sphere by \cite[Lemma 4.9]{Hempel}.
The boundary may then be filled in with a 3-ball, and the metric extended to this domain to obtain a complete, axisymmetric, simply connected Riemannian manifold $(\hat{M},\hat{g})$ with $n+1$ asymptotic ends. According to \cite{ch1,Sokolowsky} $M$ is diffeomorphic to $\mathbb{R}^3\setminus\cup_{i=1}^{n}\{p_i\}$ with the punctures $p_i$ lying on the $\hat{z}$-axis of a global system of cylindrical (Brill) coordinates $(\hat{\rho},\hat{z},\phi)$ in which $\hat{g}$ has the structure \eqref{metric}. The orbit space $M/U(1)$ is identified with the $\hat{\rho}\hat{z}$-half plane, and may be doubled across the axis so that $(\hat{\rho},\hat{z})$ parameterize $\mathbb{R}^2$ minus the axis punctures. Within this plane the projection of $\partial M$ is given by a smooth closed curve $\gamma$ which intersects the $\hat{z}$-axis at two points, and bounds a disc. Using the Riemann mapping theorem, a conformal transformation of the plane may now be applied which maps $\gamma$ to a circle centered at the origin of radius $m_1/2$. The new coordinates obtained from this map are the desired pseudospherical coordinates $(\rho,z,\phi)$. Although the punctures may move under this transformation, they will remain on the axis since the mapping is axisymmetric. Lastly, the conformal property of the map preserves the metric structure \eqref{metric}.
\end{proof}

Simple connectivity of $M$ yields potentials $v$, $\chi$, and $\psi$ satisfying \eqref{empotential}, \eqref{twistpotential} as well as the asymptotics \eqref{asym}, \eqref{asym2} in the asymptotically flat ends. In asymptotically cylindrical ends \cite{KhuriWeinstein}
\begin{equation}\label{asymcyl}
|\omega|=\rho^{2}O(r_i^{\lambda-5}),\text{ }\text{ }\text{ }\text{ }|\nabla\chi|+|\nabla\psi|=\rho O(r_i^{\lambda-3})\text{
}\text{ }\text{ as }\text{ }\text{ }r_i\rightarrow 0.
\end{equation}
The punctures $\{p_i\}_{i=1}^n$ and ball $B_{m_1/2}$ break up the $z$-axis into a sequence of connected component intervals $\{I_j\}_{j=1}^{n+2}$ on which each of the potentials is constant; this is sometimes referred to as a `rod structure'. As in \eqref{potentialcharge}, the difference of two such constants associated to the intervals surrounding a puncture yields the angular momentum or charge associated to the black hole represented by the puncture.

\begin{figure}
\includegraphics[width=13cm]{circle6_21.pdf}
\caption{Doubling with additional ends}  \label{fig}
\end{figure}

Following Section \ref{sec2} we may double the manifold and potentials across the sphere $S_{m_1/2}$ to obtain a manifold $(\bar{M},\bar{g})$ and functions $\bar{v}$, $\bar{\chi}$, $\bar{\psi}$. The only difference that occurs concerns the number of asymptotic ends. Previously the new manifold had two asymptotically flat ends, however now $\bar{M}$ has $2n+2$ asymptotic ends. In Figure \ref{fig} a diagram of the doubling rod structure in the orbit space is shown, where a single additional end occurs below the circle of radius $m_1/2$ at point $p_1$. After doubling, this end is reflected inside the circle to point $\bar{p}_1$ which represents another end of the same asymptotic type. As before the origin also becomes an asymptotically flat end, associated with the designated asymptotically flat end at infinity. Notice, as is shown in the diagram, that the potential constants on the axis also reflect inside in such a way that they are smooth across the sphere $S_{m_1/2}$ and so that the angular momentum and charge of each end inside the $B_{m_1/2}$ has the same value with opposite sign as that associated with the corresponding puncture outside the ball. Therefore the total angular momentum and total charge of $\bar{M}$, computed by adding all the individual contributions of each end, agrees with the total angular momentum and total charge of $M$.

The presence of additional asymptotically flat and cylindrical ends does not affect the proof of Lemma \ref{lemma1}, as well as scalar curvature lower bounds \eqref{scalar1} and \eqref{scalarfinal}. It follows that as before
\begin{equation}
m\geq\mathcal{I}(\Psi).
\end{equation}

\begin{proof}[Proof of Theorem \ref{thm2}]
Consider case $(ii)$. It remains to show that the renormalized energy may be minimized by a harmonic map
\begin{equation}\label{879}
\mathcal{I}(\Psi)\geq\mathcal{I}(\Psi_1).
\end{equation}
Here $\Psi_1$ is the unique renormalized harmonic map from $\mathbb{R}^3\setminus\{\text{$z$-axis}\}\rightarrow \mathbb{H}^{2}_{\mathbb{C}}$, having the same potential constants as $\Psi$. Such solutions have been constructed in \cite{KhuriWeinstein}, and the corresponding gap bound \eqref{gapbound} was established there as well. Thus, by setting $\mathcal{I}(\Psi_1)=\mathcal{F}$ we obtain the desired result.

Consider now case $(i)$. As in the proof of Theorem \ref{massam} $(i)$ let $M_0$ denote the region exterior to the outermost minimal surface $\Sigma$, with respect to the designated asymptotically flat end. Then $M_0$ is diffeomorphic to the complement of a finite number of open 3-balls and a finite number of points in $\mathbb{R}^3$, where the point punctures represent asymptotically cylindrical ends and the boundary of the 3-balls are the components of $\Sigma$. By assumption, at most one component $\Sigma_1$ of the outermost minimal surface encloses components of the boundary $\partial M$ or nonsimply connected domains. If $\Sigma_1=\emptyset$ then $M$ is simply connected and has no boundary, and therefore this theorem follows from \cite{KhuriWeinstein}. If $\Sigma_1\neq\emptyset$ let $M_1$ denote the region of $M$ outside of $\Sigma_1$. The hypotheses then imply that $M_1$ has a single component minimal surface boundary, is simply connected, and has a finite number of additional asymptotically flat and cylindrical ends. By Proposition \ref{axi} $\partial M_1=\Sigma_1$ is axisymmetric, so that $M_1$ itself is axisymmetric. The initial data $(M_1,g,k,E,B)$ now satisfy the hypotheses of Theorem \ref{thm2} $(ii)$, and \eqref{3.1} follows.
\end{proof}


TITLE: Is the convex conjugate of a strictly convex function still strictly convex?
QUESTION [4 upvotes]: Let $f:\mathbb{R}^n \to \mathbb{R}$ be a strictly convex function, i.e., for any $\lambda\in(0,1)$ and $x\neq y \in \mathbb{R}^n$, $f(\lambda x + (1-\lambda) y) < \lambda f(x) + (1-\lambda) f(y)$. Let $f^\ast$ be the convex conjugate of $f$, i.e., $f^\ast(x^\ast) = \max_{\mathbb{R}^n} \Big\{{x^\ast}^\top x - f(x)\Big\}$. Is $f^\ast$ still a strictly convex function on its domain?

REPLY [4 votes]: The answer is NO.
Let $C$ be a compact convex set containing more than just one point.
Set
$$f(x)=\sigma_C(x)+\tfrac{1}{2}\|x\|^2,$$
where $\sigma_C$ is the support function of $C$.
Note that $\sigma_C$ is convex and has full domain.
Hence
$$f\; \text{is strictly convex}$$
because $\tfrac{1}{2}\|x\|^2$ is strictly convex.
On the other hand,
$$f^* = \tfrac{1}{2}d_C^2,$$
where $d_C$ is the distance function of the set $C$.
Note that $f^*(c)=0$ for all $c\in C$, so $f^*$ is not strictly
convex because $C$ contains more than one point.


text:
{\bf Problem.} Simplify: $$\sqrt[3]{2744000}$$
{\bf Level.} Level 2
{\bf Type.} Algebra
{\bf Solution.} First, we can factor out $1000=10^3$, so $\sqrt[3]{2744000}=10\sqrt[3]{2744}$.  To proceed, we can take out factors of two from 2744, to find that $2744=2\cdot1372=2\cdot2\cdot686=2^3\cdot343$.  With a little bit of guess and check, notice that $7^3=343$.  Therefore, $10\sqrt[3]{2744}=10(14)=\boxed{140}$.

set_name: MATH


TITLE: Pasch axiom and Pythagorean field condition?
QUESTION [1 upvotes]: I am looking for a reference for the claim that the Pasch axiom is equivalent to the Pythagorean field condition, and with respect to what base theory this should be true.
Since posting the question, somebody directed me to
Szmielew, Wanda. The role of the Pasch axiom in the foundations of Euclidean geometry.  Proceedings of the Tarski Symposium (Proc. Sympos. Pure Math., Vol. XXV, Univ. California, Berkeley, Calif., 1971),  Amer. Math. Soc., Providence, R.I., 1974, pp. 123–132.
as being the source, but I don't have a copy of this yet.
Note. The reason the rationals are not enough is because the hypothenuse of a right-angle triangle with sides $1$ and $a$ needs to be in the base field in order to be able to do Euclidean geometry over it.

REPLY [8 votes]: WANDA SZMIELEW's article consider Tarski's axiom system for geometry, exposed into: What is elementary geometry ? (1957/58).

[Tarski] axiom system consists only of twelve short first order $\Pi \Sigma$ sentences and the continuity axiom, in case of the second order geometry, or the continuity axiom schema, in case of the first order (i.e., elementary) geometry.
The first version from 1930, published by Tarski only in 1948 as a note in the monograph A decision method for elementary algebra and geometry, still consists of eighteen elementary axioms, but six of them turned out step by step to be superfluous before Tarski's paper was published. After that two more axioms (identity and connectivity axioms for betweenness) were proved to depend on the remaining ones, and then the only ten left are: transitivity axiom for betweenness; Pasch's axiom:
$P: \ \ \exists a'[B(apa') \land B(ba'c)] \to \exists b'[B(bpb') \land B(ab'c)];$
identity, reflexivity, and transitivity axioms for equidistance; axiom of segment construction; five segments axiom; two dimension axioms (lower and upper); and Euclid's axiom. [...] After such a change let us denote the ten axioms by $A_1,\ldots, A10$.
The Pasch axiom is independent of all the remaining axioms together with the continuity axiom. [...] The independence model is constructed as follows: Let
$$\mathcal R=(R,+,\times,R^+)$$
be the ordered field of real numbers, $R^+$ being the positive cone of $\mathcal R$. Let the function $f$ be any one of the discontinuous one-to-one mappings of $R$ onto itself which satisfies the condition
$$f(x + y)=f(x)+f(y).$$
Then the Cartesian plane constructed over the so-called semi-ordered field
$$\mathcal R_f=(R,+,\times,f(R^+))$$
of reals is the independence model for the Pasch axiom.
Let us notice that the set $P = f(R^+)$, called the semipositive cone of $\mathcal R_f$ satisfies the conditions

(i) $x \in P \lor -x \in P$,
(ii) $x \in P \land -x \in P \to x = 0$,
(iii) $x \in P \land y \in P \to x + y \in P$,

but does not satisfy the condition

(iv) $x \in P \land y \in P \to x \times y \in P$.

In general, given a field $(F, +, \times)$ and a subset $P$ of $F$, we refer to the algebraic system
$$\mathcal F=(F,+,\times,P)$$
as a semi-ordered field iff the set $P$ (the semi-positive cone of $\mathcal F$) satisfies the conditions (i), (ii), (iii) (but not necessarily (iv)). In other words, a semi-ordered field is a field with the additive group ordered.
Since the Pasch axiom turns out to be independent, the next question arises:
  What is Euclidean geometry without the Pasch axiom? The answer is given by
  Szczerba and Szmielew in [On the Euclidean geometry without the Pasch axiom, Bull.Acad.Polon.Sci.Sér.Sci.Math.Astronom.Phys.18 (1970), 659-666, MR43#2596.].
Let us denote by $\mathcal E^P$ and call weak Euclidean geometry the theory based upon the axioms $A_1,\ldots, A10$. Now, let us drop the Pasch axiom ($A10$), but instead let us add to the remaining nine axioms two linear statements:
$A10_1: \ \ B(abc) \to B(cba)$
$A10_2: \ \ \exists b \ [B(abc) \land D(abbc)]$,
in the proofs of which, in Tarski's system, the Pasch axiom is involved. $A10_1$ is the commutativity axiom for betweenness; $A10_2$ is the axiom of midpoint existence.
Let us denote by $\mathcal E$ and call weak Euclidean Pasch-free geometry the theory based upon the eleven axioms just listed.
It is well known that $\mathcal E^P$ is the theory of Cartesian planes over arbitrary ordered (thus formally real as well) Pythagorean fields; in other words, the models of $\mathcal E^P$ coincide up to isomorphism with the Cartesian planes over such fields. Now $\mathcal E$ turns out to be the theory of Cartesian planes over arbitrary semi-ordered formally real Pythagorean fields.
Moreover, in theory $\mathcal E$ the Pasch axiom occurs to be just equivalent to the condition (iv), stating that in the semi-ordered scalar field, constructed on an arbitrarily chosen line, the semi-positive cone is closed under multiplication.
In turn, let us extend both the theories $\mathcal E^P$ and  $\mathcal E$ by the continuity axiom $Co$. In this way we get full Euclidean geometry
  $(\mathcal E^P)^{Co}$ and Euclidean Pasch-free geometry $\mathcal E^{Co}$
  both second order theories. As is shown by Szczerba in [The Pasch-free geometry with the continuity axiom, Bull.Acad.Polon.Sci.Sér.Sci.Math.Astronom.Phys.19 (1971), 613-616], the continuity axiom loses a lot of its power when the Pasch axiom is lacking. In fact, $(\mathcal E^P)^{Co}$ is known to be the theory of the Cartesian plane over the field $\mathcal R$ of real numbers, but $\mathcal E^{Co}$ turns out to be the theory of Cartesian planes not only over continuously semi-ordered fields of real numbers, but over continuously semi-ordered arbitrary formally real and Pythagorean fields. Any such field can be obtained up to  isomorphism from the field $\mathcal R$ by means of a suitable modification $\circ$ of multiplication, and thus has the power of continuum.
Conversely, every field of characteristic $0$ and power of the continuum may be
  continuously semi-ordered (characteristic $0$ is a necessary condition for a field to be semi-orderable). In conclusion, with regards to the scalar field (formally real and Pythagorean) treated as a reduct of the semi-ordered scalar field (constructed on a fixed line), the continuity axiom without the Pasch axiom determines only that its power is that of the continuum.
Finally, let us extend the theories $\mathcal E^P$  and $\mathcal E$ , instead by the continuity axiom, only by the circle axiom
$C: \ \ B(abc) \to \exists c'[B(pbc') \land D(acac')]$
This axiom says that a half-line with the origin $b$ inside or on a circle cuts this circle in a point $c'$.
$(\mathcal E^P)^C$, called geometry of elementary constructions, is known to be the theory of Cartesian planes over arbitrary ordered Euclidean fields. What is the theory $\mathcal E^C$ ? One might expect that with $\mathcal E^C$ the class of the semi-ordered formally real Euclidean fields is connected. No; it is not the case. The theories $(\mathcal E^P)^C$ and $\mathcal E^C$ turn out just to coincide with each other
$$(\mathcal E^P)^C=\mathcal E^C$$
since, as is shown by Szmielew in [The Pasch axiom as a consequence of the circle axiom, Bull.Acad.Polon.Sci.Sér.Sci.Math.Astronom.Phys.18 (1970), 751-758], the Pasch axiom is a theorem of $\mathcal E^C$. In other words, the Pasch axiom is dependent (thus superfluous) in the axiom system of geometry of elementary constructions. At first glance the situation seems to be
  paradoxical, since it is common to say that the circle axiom is a particular
  consequence of the continuity axiom, but, at the same time,
$$P \in \mathcal E^C \ \text {and} \ P \notin \mathcal E^{Co}.$$
In fact, in $(\mathcal E^P)^{Co}$, $C$ can not be proved without the use of $Co$, however in the proof of $C$ besides $Co$ also some other axioms are involved ; now it is clear that one of them must coincide with $P$.
In conclusion, the circle axiom, known to restrict the class of fields underlying the models from Pythagorean to Euclidean, in fact restricts also the semi-ordered fields to ordered ones.


TITLE: CHECK: For each of the following permutations $\rho$ in $S(6)$ write $\rho$ as the product of disjoint cycles:
QUESTION [0 upvotes]: For each of the following permutations $\rho$ in $S(6)$ write $\rho$ as the product of disjoint cycles:

$\begin{pmatrix} 1 & 5 & 2 \end{pmatrix}^{-1} \begin{pmatrix} 2 & 4
    & 5 & 6\end{pmatrix}\begin{pmatrix} 1 & 6 & 2 \end{pmatrix}^{-1}$
$[\begin{pmatrix} 3 & 6 & 5 & 4 \end{pmatrix}\begin{pmatrix} 1 & 4 &
    3 & 6 \end{pmatrix}]^{-1}[\begin{pmatrix} 1 & 6
    \end{pmatrix}\begin{pmatrix} 2 & 3 \end{pmatrix}\begin{pmatrix} 1 &
    4 \end{pmatrix}]^{-1}$

Part 1:
\begin{equation*}
\begin{split}
& \begin{pmatrix} 1 & 5 & 2 \end{pmatrix}^{-1} \begin{pmatrix} 2 & 4 & 5 & 6\end{pmatrix}\begin{pmatrix} 1 & 6 & 2 \end{pmatrix}^{-1}
= \begin{pmatrix} 2 & 5 & 1 \end{pmatrix} \begin{pmatrix} 2 & 4 & 5 & 6 \end{pmatrix}\begin{pmatrix} 2 & 6 & 1 \end{pmatrix} \\
= & \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 5 & 3 & 1 & 6 & 2 \end{pmatrix}=\begin{pmatrix} 1 & 4 \end{pmatrix}\begin{pmatrix} 2 & 5 & 6 \end{pmatrix}\begin{pmatrix} 3 \end{pmatrix}
= \begin{pmatrix} 1 & 4 \end{pmatrix}\begin{pmatrix} 2 & 5 & 6 \end{pmatrix} \\
\end{split}
\end{equation*}
Part 2:
\begin{equation*}
\begin{split}
& [\begin{pmatrix} 3 & 6 & 5 & 4 \end{pmatrix}\begin{pmatrix} 1 & 4 & 3 & 6 \end{pmatrix}]^{-1}[\begin{pmatrix} 1 & 6 \end{pmatrix}\begin{pmatrix} 2 & 3 \end{pmatrix}\begin{pmatrix} 1 & 4 \end{pmatrix}]^{-1} \\
= & \begin{pmatrix} 4 & 5 & 6 & 3 \end{pmatrix} \begin{pmatrix} 6 & 3 & 4 & 1 \end{pmatrix}\begin{pmatrix} 1 & 6 \end{pmatrix}\begin{pmatrix} 2 & 3 \end{pmatrix}\begin{pmatrix} 1 & 4 \end{pmatrix} \\
= & \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 5 & 2 & 3 & 6 & 1 \end{pmatrix}
= \begin{pmatrix} 1 & 4 & 3 & 2 & 5 & 6 \end{pmatrix}
\end{split}
\end{equation*}
attempt 2:
\begin{equation*}
\begin{split}
& [\begin{pmatrix} 3 & 6 & 5 & 4 \end{pmatrix}\begin{pmatrix} 1 & 4 & 3 & 6 \end{pmatrix}]^{-1}[\begin{pmatrix} 1 & 6 \end{pmatrix}\begin{pmatrix} 2 & 3 \end{pmatrix}\begin{pmatrix} 1 & 4 \end{pmatrix}]^{-1} \\
= & [\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 5 & 6 & 4 & 1   \end{pmatrix}]^{-1}[\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 3 & 2 & 6 & 5 & 1 \end{pmatrix}]^{-1} \\
= &  \begin{pmatrix} 1 & 4 & 6 & 5 & 3 \end{pmatrix}\begin{pmatrix} 1 & 6 & 2 & 3 & 4 \end{pmatrix} \\
= & \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 1 & 6 & 4 & 3 & 2 \end{pmatrix}
= \begin{pmatrix} 5 & 1 & 6 & 3 & 2 \end{pmatrix}
\end{split}
\end{equation*}
Are these correct?

REPLY [0 votes]: $\textbf{Part a:}$
\begin{equation*}
\begin{split}
& \begin{pmatrix} 1 & 5 & 2 \end{pmatrix}^{-1} \begin{pmatrix} 2 & 4 & 5 & 6\end{pmatrix}\begin{pmatrix} 1 & 6 & 2 \end{pmatrix}^{-1}
= \begin{pmatrix} 2 & 5 & 1 \end{pmatrix} \begin{pmatrix} 2 & 4 & 5 & 6 \end{pmatrix}\begin{pmatrix} 2 & 6 & 1 \end{pmatrix} \\
= & \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 5 & 3 & 1 & 6 & 2 \end{pmatrix}=\begin{pmatrix} 1 & 4 \end{pmatrix}\begin{pmatrix} 2 & 5 & 6 \end{pmatrix}\begin{pmatrix} 3 \end{pmatrix}
= \begin{pmatrix} 1 & 4 \end{pmatrix}\begin{pmatrix} 2 & 5 & 6 \end{pmatrix} \\
\end{split}
\end{equation*}
$\textbf{Part b:}$
\begin{equation*}
\begin{split}
& [\begin{pmatrix} 3 & 6 & 5 & 4 \end{pmatrix}\begin{pmatrix} 1 & 4 & 3 & 6 \end{pmatrix}]^{-1}[\begin{pmatrix} 1 & 6 \end{pmatrix}\begin{pmatrix} 2 & 3 \end{pmatrix}\begin{pmatrix} 1 & 4 \end{pmatrix}]^{-1} \\
= & [\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 5 & 6 & 4 & 1 \end{pmatrix}]^{-1}[\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 3 & 2 & 6 & 5 & 1 \end{pmatrix}]^{-1} \\
= & [\begin{pmatrix} 1 & 3 & 5 & 4 & 6 \end{pmatrix}]^{-1}[\begin{pmatrix} 1 & 4 & 6 \end{pmatrix}\begin{pmatrix} 2 & 3 \end{pmatrix}]^{-1} \\
= & \begin{pmatrix} 6 & 4 & 5 & 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 3 \end{pmatrix}\begin{pmatrix} 6 & 4 & 1 \end{pmatrix} \\
= & \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 1 & 2 & 6 & 3 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 4 & 6 & 5 & 3 & 2 \end{pmatrix} \\
\end{split}
\end{equation*}


TITLE: Interesting question regarding evolution of algorithm under certain conditions
QUESTION [1 upvotes]: On a random string of six digits containing numbers of the set $S=\{1,2,3,4,5,6\}$, repeat the following operation:
If $k$ is the first number of the string, then reverse the order of first $k$ numbers of the string.
For example: $342561\rightarrow 243561\rightarrow 423561\rightarrow 532461\rightarrow 642351\rightarrow 153426$
Prove that any such string would terminate with $1$ at the first position.
Now on checking a few examples, I have made a few observations:
$1.$ The last digit of the string is an increasing monovariant.
$2.$ The first digit of string is always a different number unless the last digit changes. This means that until the last digit of the string change, no number that appeared in the first position in the string will ever repeat.
$3.$ When the last digit reaches its maximum value (which is $6$ here), the observation $2$ works the same but for th second last digit now instead of the last one.
Now the first observation can be explained by the fact that the last digit will only change when we would have $6$ in the first position and no number in the string is greater than $6$.
The third observation can be explained by the fact that when the last digit reaches the value of $6$, this value would never change as all the numbers are less than $6$. Thus, the observation $2$ would now work with the second last digit instead of the last digit since it would be the same as applying the algorithm to  a string of length $5$ instead of $6$.
Now if I can prove observation $2$, then I can say that since no number occuring in the first position repeats, we would have either $6$ or $1$ in the first position sometime in atmost $6$ steps. If we obtain $6$ before $1$ in the first position, then $6$ would move to the last position. Similarily after $6$ have moved to the last position, we would either have $1$ or $5$ in the first position after atmost $5$ steps and so on.
This indicates that after some finite steps, the sequence is bound to have $1$ at the first position.
Now I need help to prove observation $2$. Please help.
THANKS

REPLY [1 votes]: Let us consider a generalized version of this problem where we are given some permutation $\pi$ of $[1, n]$ and let $f$ denote the function mapping such a permutation to the one resulting by applying the given operation.
Further, we will call $\pi$ correct for $k$ if $\pi(k) = k$.
We begin by observing 2 properties of $f$:

$f$ preserves the correctness of a permutation $\pi$ at $k$ if $\pi(1) < k$, i.e. reversing the order of the first $\ell < k$ elements of the permutation does not change the position of $k$.
$f(\pi)$ is correct for $\pi(1)$, i.e. $f(\pi)$ is correct for the first element of $\pi$.

A simple corollary of these properties is that if $\pi$ is correct at $k$ and $f(\pi)$ is not then $f(\pi)$ must be correct for some $j > k$ on which $\pi$ is not correct.
This follows from the fact that $k$ can only be moved to an incorrect position if the first element of $\pi$ is some $j > k$.
Now consider the following energy function $E$:
$$ E \colon S_{[1, n]} \to \mathbb R,
\pi \mapsto \sum_{\pi \text{ correct for } k} 2^{k} $$
Using the properties of $f$ established above we find that $E(f(\pi)) \geq E(\pi)$ with the inequality being strict exactly for all $\pi$ on which $f$ is not stationary.
As $S_{[1, n]}$ is finite, it follows that the state $f^m(\pi)$ of the algorithm (started on some initial permutation $\pi$) attains a maximal energy value after $p \leq |S_{[1, n]}| < \infty$ steps. We conclude that $f^p(\pi)$ must hence be a stationary point of $f$ which is equivalent to $f^p(\pi)$ being correct for 1.


TITLE: if $a=\infty$ It can be concluded that ?: $\int_{-\infty}^{\infty}f(x)=0$
QUESTION [1 upvotes]: I know that if $f$ be odd function and $a,-a \in D_f$ Then :
$$\int_{-a}^af(x)=0$$
Now if $a=\infty$ It can be concluded that ?:
$$\int_{-\infty}^{\infty}f(x)=0$$

REPLY [5 votes]: No, the improper integral $\displaystyle\int_{-\infty}^{\infty}xdx$ does not exist.


TITLE: Why is it clear that $X_{n+1}$ is independent of $\sigma (X_{1},...,X_{n})$ for IID random variables
QUESTION [2 upvotes]: Why is it clear that $X_{n+1}$ is independent of $\sigma (X_{1},...,X_{n})$
Example:
Let $(X_{n})_{n}$ be a sequence of IID RV's where $E[X_{1}]=0$, now define $S_{n}:=\sum\limits_{i=1}^{n}X_{i}$
Let $\mathcal{F}_{n}:=\sigma (X_{1},...,X_{n})$ and show $E[S_{n+1}\vert \mathcal{F}_{n}]=S_{n}$
I agree with the solution up to a point:
$E[S_{n+1}\vert \mathcal{F}_{n}]=\sum\limits_{i=1}^{n+1}E[X_{i}\vert \mathcal{F}_{n}]=E[X_{n+1}\vert \mathcal{F}_{n}]+\sum\limits_{i=1}^{n}X_{i}$
and then it is said that $X_{n+1}$ is independent of $\mathcal{F}_{n}$, and I do not agree with that unless I do not understand IID correctly. To my understanding, IID in this case would mean:
$X_{n+1}$ is independent from $\sigma (X_{i})$ where $1\leq i\leq n$. The definition does not say that $X_{n+1}$ is independent of $\sigma (X_{1},...,X_{n})$. Or is that in actual fact the definition of IID?

REPLY [2 votes]: The first "i" in "iid" means that the family of random variables $(X_i)_{i\in\mathbb N}$ is mutually independent. By definition, this means that for every pair of disjoint sets $A,B\subseteq \mathbb N$ the $\sigma$-algebras $\sigma(X_i\colon i\in A)$ and $\sigma(X_j\colon j\in B)$ are independent. In particular, taking $A=\{n+1\}$ and $B=\{1,\ldots,n\}$ answers your question.


text:
\subsection{Proving non-rationalizability of a set of square roots}
\label{sec:Alphabets}

In the context of Feynman integral computations, it is often not enough to study the rationalizability of a single square root.
Instead, practitioners are usually interested in whether or not several different square roots can be rationalized simultaneously.
However, we will see that the non-rationalizability of a set of square roots (also called \textit{alphabet})
can often be deduced from the non-rationalizability of a single square root so that many of our previous methods can also be applied in this more general context.

As before, we fix $k=\mathbb{C}$,
and write $R=k[X_1,...,X_n]$ for the ring of polynomials and $Q=\mathrm{Frac}\;R$ for its field of fractions.

\begin{definition}
Let $f_1,...,f_r$ be polynomials in $R$.
An alphabet $\{ \sqrt{f_1},...,\sqrt{f_r}\}$ is called {\em rationalizable}
if there is a homomorphism of $k$-algebras $\phi\colon Q\to Q$ such that $\; \phi(f_i)=h_i^2$ for some $h_i\in Q$, where $i=1,...,r$.
\end{definition}

\begin{remark}
From the definition it immediately follows that if an alphabet $\mathcal{A}^\prime$ is non-ra\-tio\-na\-li\-za\-ble,
then every alphabet $\mathcal{A}\supseteq \mathcal{A}^\prime$ containing $\mathcal{A}^\prime$ is also non-rationalizable.

This remark is particularly useful when $\mathcal{A}$ has a subset $\mathcal{A}^\prime$ containing only square roots of polynomials in fewer variables, that is, after possibly reordering the variables and the polynomials,
$$
\mathcal{A}^\prime=\Big\{\sqrt{f_1},...,\sqrt{f_s}\Big\}
$$
with $s<r$ and $f_1,...,f_s\in k[X_1,...,X_m]\subset R$, $m<n$.
Then one can try to disprove the rationalizability of $\mathcal{A}$ by disproving the rationalizability of $\mathcal{A}^\prime$.
The rationalizability of $\mathcal{A}^\prime$ as alphabet of square roots of polynomials in $R$ is equivalent to its rationalizability as square roots of polynomials in $k[X_1,...,X_m]$ (Corollary~\ref{c:Cones}).
The latter task is easier because of the fewer variables involved and, in particular, if $m=1,2$, one can then apply the criteria given in Subsections~\ref{sec:OneVariable} and~\ref{sec:TwoVariables}.
\end{remark}

\begin{proposition}\label{p:MoreRoots}
If the alphabet $\{ \sqrt{f_1},...,\sqrt{f_n}\}$ is rationalizable then, for every non-empty subset $J\subseteq \{1,...,n\}$, the square root
\begin{equation}\label{eq:productRoot}
\sqrt{\prod_{j\in J} f_j}
\end{equation}
is rationalizable.
\end{proposition}
\begin{proof}
By definition, if $\{ \sqrt{f_1},...,\sqrt{f_n}\}$ is rationalizable, then there exists a non-constant $k$-algebra homomorphism $\phi\colon Q\to Q$ such that, for $i=1,...,n$, the map $\phi$ sends $f_i$ to $h_i^2$, for some $h_i\in Q$.
In particular, $\phi(f_j)=h_j^2$ for every $j\in J$.
Hence, $\phi (\prod_{j \in J}f_j)=(\prod_{j\in J}h_j)^2$,
proving the statement.
\end{proof}
\begin{remark}\label{r:AlphabetStrategy}
The above straightforward proposition allows us to prove that a given alphabet is not rationalizable by showing that at least one square root of the form \eqref{eq:productRoot} is not rationalizable.

At the moment, we do not know whether the converse statement of Proposition~\ref{p:MoreRoots} holds or not,
as even if the product $\sqrt{\prod_{j\in J} f_j}$ is rationalizable for every $J\subseteq \{1,...,n\}$,
the rationalizing morphisms $\phi_J$ do not need to be a priori all equal.
Nevertheless, one can often prove the rationalizability of a given alphabet by providing an explicit variable change that rationalizes all of its square roots simultaneously.
To find such a variable change, one can try to apply the elementary strategy mentioned in Remark~\ref{r:StartingPoint}, which works for many practical examples.
\end{remark}

Let us apply Proposition~\ref{p:MoreRoots} to the alphabets of some recent Feynman integral computations.
\begin{example}
By Corollary~\ref{c:DegreeCriterion}, square roots of a squarefree polynomial in one variable of degree $d>2$ are not rationalizable.
Such square roots occurred in many Feynman integral computations of the last decade, see  \cite{Laporta:2004rb,MullerStach:2011ru,Adams:2013kgc,Bloch:2013tra,Adams:2014vja,Adams:2015gva,Adams:2015ydq,Sogaard:2014jla,Bloch:2016izu,Remiddi:2016gno,Adams:2016xah,Bonciani:2016qxi,vonManteuffel:2017hms,Adams:2017ejb,Bogner:2017vim,Ablinger:2017bjx,Remiddi:2017har,Bourjaily:2017bsb,Hidding:2017jkk,Broedel:2017kkb,Broedel:2017siw,Broedel:2018iwv,Adams:2018yfj,Adams:2018bsn,Adams:2018kez}.
As an example, let us consider the following alphabet, which appears in perturbative corrections for Higgs production in \cite{Dulat:talk2018, Anastasiou:2015yha}:
\begin{equation*}
    {\mathcal A}=
 \Big\{
 \sqrt{X},
 \sqrt{1+4X},
 \sqrt{X (X-4)}\Big\}.
\end{equation*}

These three square roots cannot be rationalized by a single rational variable change.
To see this, define $f_1:=X, f_2:=1+4X, f_3:=X(X-4)$,
take $J:=\{2,3\}$, and consider the square root
\begin{equation}\label{eq:higgsRoot}
\sqrt{\prod_{j\in J}f_j}=\sqrt{(1+4X)X(X-4)}.
\end{equation}
Note that the product $(1+4X)X(X-4)$ is a squarefree polynomial of degree $3>2$.
Therefore, by Corollary~\ref{c:DegreeCriterion}, the square root \eqref{eq:higgsRoot} is not rationalizable.
Thus, by Proposition~\ref{p:MoreRoots}, we conclude that $\mathcal{A}$ is not  rationalizable.
\end{example}

\begin{example}\label{e:dijetproduction}
Let us now consider the following set of square roots that is relevant for perturbative corrections to di-photon and di-jet hadro-production in \cite{Becchetti:2017abb}:

\begin{align*}
    \mathcal{A}=&\Big\{\; \sqrt{X+1},\; \sqrt{X-1}, \; \sqrt{Y+1},\\  & \sqrt{X+Y+1},\; \sqrt{16X+(4+Y)^2}\; \Big\}.
\end{align*}
Write $f_1,..., f_5$ for the polynomial arguments of the square roots in $\mathcal{A}$.
To show that $\mathcal{A}$ is not  rationalizable, consider the whole set of indices $J=\{1, 2, 3, 4, 5\}$
and define
$$
f:=\prod_{j\in J} f_j=(X+1)(X-1)(Y+1)(X+Y+1)(16X+(4+Y)^2).
$$
Let $\overline{S}$ be the associated double cover of $\sqrt{f}$ (cf. Definition~\ref{def:AssDoubleCover}).
It is easy to check---for example by using our {\tt Magma} function---that $\overline{S}$ has only rational simple singularities.
Since $f$ has degree $6$,
Theorem~\ref{thm:DegreeCriterion} tells us that $\sqrt{f}$ is not rationalizable.
Hence,
using Proposition~\ref{p:MoreRoots},
we can conclude that $\mathcal{A}$ cannot be  rationalizable.
\end{example}

\begin{example}
It is important (and fair) to stress that the results presented in this paper are not always enough to get an answer.
Consider the alphabet
\begin{align*}
    \mathcal{A}=\Big\{ &\sqrt{X_1(X_1-4X_3)}, \sqrt{-X_1X_2(4X_3(X_3+X_2)-X_1X_2)},\\
    &\sqrt{X_1(X_2^2(X_1-4X_3)+X_3X_1(X_3-2X_2))}\,  \Big\}
\end{align*}
 relevant for two-loop EW-QCD corrections to Drell--Yan scattering (cf.~\cite{Heller:2019gkq,Besier:2019hqd,Bonciani:2016ypc}).
Denote by $F_1, F_2, F_3$ the polynomial arguments of the square roots in $\mathcal{A}$.
Proving non-rationalizability of~$\mathcal{A}$ requires more than just the techniques presented in this paper.
Notice that $F_1, F_2, F_3$ are all homogeneous.
Therefore, we can view them as the homogenizations of three polynomials with respect to one of the three variables, for example $X_3$.
Studying the rationalizability of $\mathcal{A}$ is, hence, equivalent to studying the rationalizability of
$$
\Big\{\; \sqrt{f_1}, \sqrt{f_2}, \sqrt{f_3}\; \Big\},
$$
where $f_i=f_i(X,Y)$ is the dehomogenization of $F_i$ with respect to $X_3$, that is,
$$
f_i(X,Y):=F_i(X,Y,1).
$$

As $f_1, f_2$ and $f_3$ have degree $2, 4$ and $4$ respectively, and since their associated double covers have at most rational simple singularities, one has that $\sqrt{f_1},\sqrt{f_2},\sqrt{f_3}$ are rationalizable when considered individually (cf. Theorem~\ref{thm:DegreeCriterion}).
The products $f_1f_2$ and $f_1f_3$, after removing the square factors, also have degree $4$ and associated double covers with at most rational simple singularities.
Hence, their square roots are rationalizable.
The product $f_2f_3$ has,
after removing square factors, degree $6$ but its associated double cover has (at least) two non-simple singularities, so we cannot conclude that $\sqrt{f_2f_3}$ is not rationalizable.
(After further investigation it turns out that it is in fact rationalizable.)
We are left with the square root of the product $f_1f_2f_3$.
After removing the squares,
the product has degree $8$ but the associated double cover has some non-simple singularities as well.
So again, we cannot use Theorem~\ref{thm:DegreeCriterion} to conclude that its square root is non-rationalizable.

Analogous computations and results are obtained if one dehomogenizes the polynomials $F_1,F_2,F_3$ with respect to $X_1$ or $X_2$.
For this reason, using the results and techniques of the previous subsections, we cannot prove or disprove the rationalizability of the alphabet $\mathcal{A}$.

Only by using methods that are beyond the scope of this paper, one can see that
the square root $\sqrt{f_1f_2f_3}$ is not rationalizable and hence that the alphabet $\mathcal{A}$ is not rationalizable, where the $f_i$ denote the above mentioned dehomogenizations with respect to $X_3$.

\end{example}

\begin{remark}\label{r:StartingPoint}
Finally, let us stress that,
when trying to prove non-rationalizability in physics computations,
it is crucial to pick the right starting point for the proof.
To clarify this important subtlety, consider the following alphabet:
\begin{equation*}\label{eq:initsetstartingpoint}
    \mathcal{A}:=\left\{\sqrt{X-1},\sqrt{X-2}\right\}.
\end{equation*}
To rationalize this set, we could proceed as follows:
\begin{enumerate}
    \item try to rationalize the first square root;
    \item if successful, plug the corresponding substitution into the second square root and try to rationalize the resulting square root;
    \item if successful, compose both substitutions to obtain a single substitution that will rationalize both square roots.
\end{enumerate}
(A more detailed discussion of this procedure can be found in \cite{Besier:2019kco, Besier:2018jen}.)
We start out with the rationalization of the first square root, via the homomorphism $\phi\colon k(X)\to k(X)$ defined by $\phi\colon X\mapsto X^4+1$, hence $\sqrt{\phi(X-1)}=X^2$.
Using $\phi$, the second square root becomes
\begin{equation}\label{eq:nonratstartingpoint}
    \sqrt{X^4-1}\, ,
\end{equation}
giving us a non-rationalizable square root, cf. Theorem~\ref{thm:CriterionForCurves}.

Therefore, one might be tempted to assume that the non-rationalizability of \eqref{eq:nonratstartingpoint} implies non-rationalizability of $\mathcal{A}$.
This assumption is, however, not true:
consider the homomorphism $\psi\colon k(X)\to k(X)$ defined as $\psi\colon X\mapsto X^2+1$.
Then one can easily see that $\psi (X-1)=X^2$ and $\psi(X-2)=X^2-1$.
Notice that $X^2-1$  has degree $2$ and so one can easily rationalize its square root (cf. Corollary~\ref{c:SingProj}).
A suitable substitution is, for example, given by the homomorphism
$\sigma\colon k(X)\to k(X)$ defined by
$\sigma\colon X\mapsto \frac{2X^2}{1-X^2}+1$.
Finally, the composition
$$
\iota:=\sigma\circ\psi\colon k(X)\to k(X), \; X\mapsto
\frac{2(X^2+1)^2}{1-(X^2+1)^2}+1
$$
rationalizes both square roots simultaneously, proving that $\mathcal{A}$ is rationalizable.

Besides illustrating a way to prove rationalizability for an alphabet, this example gives us the following important insight:
proving non-rationalizability after some substitutions have already been made does, in general, not imply non-rationalizability of the original alphabet.
For Feynman integral computations, this means that one should always prove non-rationalizability as early as possible, i.e., as soon as the square roots arise in the computation.
\end{remark}

config: arxiv


text:
\section{Setup and assumptions}
\label{sec:setup}

\begin{table}[t]
    \centering
    \caption{Summary of notation}
    \label{table:notation-recap}
    \vskip -2em
    \null\hfill\begin{minipage}[t]{.45\textwidth}
    \centering~\\
    \begin{tabular}{ll}
        \toprule
        {\bf Concept}                            & {\bf Symbol}
        \\ \midrule
        Iteration counter, maximum         & $k$, $T$
        \\ Iterates, minimum                     & $\x_k, \x^*$
        \\ Step-size                             & $\eta_k$
        \\ Function value, minimum               & $f(\x), f^*$
        \\ Stoch. function value, minimum    & $f_i(\x), f_i^*$
        ~\vphantom{Variance}\\
        \\ \bottomrule
    \end{tabular}
    \end{minipage}\hspace{2.5em}
    \begin{minipage}[t]{.45\textwidth}
    \centering~\\
    \begin{tabular}{ll}
        \toprule
        {\bf Concept}              & {\bf Symbol}
        \\ \midrule
        General preconditioner  & $A_k$
        \\ Preconditioner bounds   & $[a_{\min}, a_{\max}]$
        \\ Maximum smoothness      & $L_{\max}$
        \\ Dimensionality          & $d$
        \\ Diameter bound          & $D$
        \\ Variance                & $\sigma^{\raisebox{-1pt}{\tiny $2$}} = \E_i[f_i(\xopt) - f^*_i]$
        \\ \bottomrule
    \end{tabular}
    \end{minipage}\hfill\null\\[1em]
\end{table}

We restate the main notation in \cref{table:notation-recap}. We now restate the main assumptions required for our theoretical results

We assume our  objective $f:\R^d\rightarrow\R$ has a finite-sum structure,
\begin{equation}
    f(\x) = \frac{1}{n}\sum_{i=1}^n f_i(\x),
\end{equation}
and analyze the following update, with $i_k$ selected uniformly at random,
\begin{equation}
    \label{eq:appendix-update-rule}
    \xkk = \xk - \etak \,  \smallAkInverse \mk  \quad \text{;} \quad \mk = \beta  \mkm + (1 - \beta) \gradk{\xk}
    \tag{Update rule}
\end{equation}
where $\etak$ is either a pre-specified constant or selected on the fly. We consider \adagrad/ and \amsgrad/ and use the fact that the preconditioners are non-decreasing i.e. $A_k \succeq A_{k-1}$. For \adagrad/, $\beta = 0$. For \amsgrad/, we further assume that the preconditioners remain bounded
with eigenvalues in the range $[a_{\min}, a_{\max}]$,
\aligns{
    a_{\min} I \preceq A_k \preceq a_{\max} I.
    \tag{Bounded preconditioner}
    \label{eq:bounded-preconditioner}
}
For all algorithms,
we assume that the iterates do not diverge and remain in  a ball of radius $D$,
as is standard in the literature on online learning~\citep{duchi2011adaptive,levy2018online} and adaptive gradient methods~\citep{reddi2019convergence},
\aligns{
    \norm{\xk - \xopt} \leq D.
    \tag{Bounded iterates}
    \label{eq:bounded-iterates}
}
Our main assumptions are that each individual function $f_i$
is convex, differentiable, has a finite minimum $f_i^*$, and is $L_i$-smooth, meaning that for all $v$ and $w$,
\aligns{
    f_i(v) &\geq f_i(w) - \lin{\nabla f_i(w), w-v},
    \tag{Individual Convexity}
    \label{eq:individual-convexity}
    \\
    f_i(v) & \leq f_i(w) + \inner{\nabla f_i(w)}{v - w} + \frac{L_i}{2} \normsq{v - w},
    \tag{Individual Smoothness}
    \label{eq:individual-smoothness}
}
which also implies that $f$ is convex and $L_{\max}$-smooth, where $L_{\max}$
is the maximum smoothness constant of the individual functions.
A consequence of smoothness is the following bound on the norm of the gradient stochastic gradients,
\aligns{
    \norm{\nabla f_i(\x)}^2
    \leq
    2 L_{\max} (f_i(\x) - f_i^*).
}
To characterize interpolation, we define the
expected difference between the minimum of $f$, $f(\xopt)$,
and the minimum of the individual functions $f_i^*$,
\aligns{
    \sigma^2 &= \Expect[i]{f_i(\xopt) - f_i^*} < \infty.
    \tag{Noise}
}
When interpolation is exactly satisfied, every data point can be fit exactly, such that $f_i^* = 0$ and $f(\xopt) = 0$, we have $\sigma^2 = 0$.

\iffalse
\begin{lemma}[Starting points]
\label{lem:starting-point}
The update $w_{k+1} = w_k - \eta_k A_k^{-1} \nabla f_{i_k}(w_k)$ leads to
\aligns{
    2 \sum_{k=1}^T \lin{\nabla f_{i_k}(w_k), w_k - w^*}
    &\leq
    \sum_{k=1}^T
    \norm{w_k - w^*}^2_{\paren{
        \frac{1}{\eta_k} A_k - \frac{1}{\eta_{k-1}} A_{k-1}
    }}
    +  \sum_{k=1}^T \eta_k \norm{\nabla f_{i_k}(w_k)}^2_{A_k^{-1}},
    \\
    2 \sum_{k=1}^T \lin{A_k^{-1} \nabla f_{i_k}(w_k), w_k - w^*}
    &\leq
    \sum_{k=1}^T
    \norm{w_k - w^*}^2_{\paren{
        \frac{1}{\eta_k} - \frac{1}{\eta_{k-1}}
    }}
    +  \sum_{k=1}^T \eta_k \norm{A_k^{-1} \nabla f_{i_k}(w_k)}^2.
}
where $\norm{\x - \xopt}_A^2 = (\x - \xopt)^\top \m! A (\x -\xopt)$ and
$\norm{\x - \xopt}^2_{\frac{1}{\eta_k} - \frac{1}{\eta_{k-1}}} = \paren{\frac{1}{\eta_k} - \frac{1}{\eta_{k-1}}}\norm{\x-\xopt}^2$.
\end{lemma}
\begin{proof}
The starting point of the analysis is the recursion
\aligns{
    \norm{w_{k+1} - w^*}^2_{A_k}
    =
    \norm{w_k - w^*}^2_{A_k} - 2 \eta_k \lin{\nabla f_{i_k}(w_k), w_k - w^*} + \eta_k^2 \norm{\nabla f_{i_k}(w_k)}^2_{A_k^{-1}},
}
Dividing by $\eta_k$ and reorganizing the terms yields
\aligns{
    2 \lin{\nabla f_{i_k}(w_k), w_k - w^*}
    =
    \frac{1}{\eta_k}\paren{
    \norm{w_k - w^*}^2_{A_k}
    - \norm{w_{k+1} - w^*}^2_{A_k}
    }
    + \eta_k \norm{\nabla f_{i_k}(w_k)}^2_{A_k^{-1}},
}
and summing $T$ iterations of the process gives
\aligns{
    2 \sum_{k=1}^T \lin{\nabla f_{i_k}(w_k), w_k - w^*}
    =
     \sum_{k=1}^T
    \frac{1}{\eta_k}\paren{
    \norm{w_k - w^*}^2_{A_k}
    - \norm{w_{k+1} - w^*}^2_{A_k}
    }
    +  \sum_{k=1}^T \eta_k \norm{\nabla f_{i_k}(w_k)}^2_{A_k^{-1}}.
}
Reorganizing the sum of iterate distances and ignore negative boundary terms,
\aligns{
    2 \sum_{k=1}^T \lin{\nabla f_{i_k}(w_k), w_k - w^*}
    \leq
    \sum_{k=1}^T
    \norm{w_k - w^*}^2_{\paren{
        \frac{1}{\eta_k} A_k - \frac{1}{\eta_{k-1}} A_{k-1}
    }}
    +  \sum_{k=1}^T \eta_k \norm{\nabla f_{i_k}(w_k)}^2_{A_k^{-1}}.
}
Applying the same steps to the Euclidean norm $\norm{\cdot}$ rather than the preconditioned norm $\norm{\cdot}_{A_k}$ yields the second inequality. \qedhere
\end{proof}
\fi


TITLE: Verifying the Gaussian Transformation of $exp\left\{\frac{1}{2}\sum_{i,j} S_i J_{ij} S_j\right\}$ from "Advanced Mean Field Methods"
QUESTION [1 upvotes]: The book Advanced Mean Field Methods mentions the following equation as a result of a "simple gaussian transformation".
$$
exp\left\{\frac{1}{2}\cdot\textbf{s}^T \cdot \textbf{J} \cdot\textbf{s}\right\} = \frac{1}{(2\pi)^{N/2}\sqrt{det(J)}} \int_{-\infty}^\infty \prod_i dx_i \exp\left\{-\frac{1}{2}\sum_{ij}x_i(\textbf{J}^{-1})_{ij}x_j + \sum_ix_iS_i\right\}
$$
It is assumed that $\textbf{J}$ is positive definite. I would like to derive this for myself as I try to with all the expression I come across. Looking at the right-hand side, it is very similar to a multivariate normal distribution with $\mu = \textbf{s}$ and $\Sigma=\textbf{J}$
$$\frac{1}{(2\pi)^{N/2}\sqrt{det(J)}} \exp\left\{-\frac{1}{2}(\textbf{x}-\textbf{s})^T\textbf{J}^{-1}(\textbf{x}-\textbf{s})\right\}$$
$$
= \frac{1}{(2\pi)^{N/2}\sqrt{det(J)}} \exp\left\{-\frac{1}{2}\left(\textbf{x}^T\textbf{J}^{-1}\textbf{x} + \textbf{s}^T\textbf{J}^{-1}\textbf{s}\right) +\textbf{x}^T\textbf{J}^{-1}\textbf{s}\right\}.
$$
We are missing some terms and gaining some extra terms. Here, I don't know how to proceed, or if I am on the right track. I would like to take the expectation of some $g(\textbf{x}) =e^{f(\textbf{x} ; \textbf{s})}$ with respect to the multivariate normal above to cancel the extra terms and add the missing ones. However, my attempts at solving for such an $g$ does not lead to an easily interpretable one nor give reason to think we would have equality. I feel I am not taking the right approach.
I would greatly appreciate ideas and insights. Thank you.

REPLY [2 votes]: Let's consider the argument of the exponential on the R.H.S.:
$$
-\frac{1}{2} \bf{x}^T \bf{J}^{-1} \bf{x} + \bf{x}^T\cdot \bf{s}
$$
the trick is to add and subtract here a suitable quantity to complete the square. The correct choice (suggested by the solution you have) is clearly
$$
\frac{1}{2}\bf{s}^T \bf{J} \bf{s}.
$$
After this your exponent reads:
$$
\left(-\frac{1}{2} \bf{x}^T \bf{J}^{-1} \bf{x} + \bf{x}^T\cdot \bf{s} - \frac{1}{2}\bf{s}^T \bf{J} \bf{s} \right) + \frac{1}{2}\bf{s}^T \bf{J} \bf{s}
$$
Now the term in brackets can be written as a quadratic form $-\frac{1}{2} \bf{y}^T \bf{J}^{-1} \bf{y}$ if you define
$$
\bf{y} = \bf{x} - \bf{J}\bf{s}
$$
and with the assumption that $\bf{J}$ is a symmetric matrix, so that $\bf{J}^T=\bf{J}$.
In conclusion, your integral can be performed with a linear change of variable $\bf{x} \to \bf{y}$ (notice that this is simply a shift of your variable, so the extrema remain infinite and the Jacobian of the transformation is the identity):
$$
\frac{\exp{\left[\frac{1}{2}\bf{s}^T \bf{J} \bf{s} \right]}}{(2\pi)^{N/2}\sqrt{\det{\bf{J}}}} \int D\bf{y} \exp{\left[-\frac{1}{2} \bf{y}^T \bf{J}^{-1} \bf{y}\right]} = \exp{\left[\frac{1}{2}\bf{s}^T \bf{J} \bf{s} \right]}.
$$
where I have used the standard Gaussian integral result in the last equality.
Notice that this formula is a very useful trick to study mean field theory in classical and quantum field theories of interacting systems, and is often called Hubbard-Stratonovich transformation.


TITLE: Prove two sets are not equal. Discrete Mathematics
QUESTION [1 upvotes]: Let $S=\{x\in\Bbb Z:\exists y\in\Bbb Z:x=24y\}$ and $T=\{x\in\Bbb Z:\exists y,z\in\Bbb Z:x=4y\land x=6z\}$. Prove that $S\ne T$.
Here's what I've done so far. I know that $S$ is a subset of $T$, but $T$ is not a subset of $S$. So $S$ cannot equal $T$. I am having trouble proving $T$ is not a subset of $S$.
suppose $x$ is in the set of $T$. Therefore $x$ is divisible by $4$ and $6$. How do I finish off this proof?
Edit: Consider $24 = 4y$ and $24 = 6z$. $y = 6$ while $z = 4$. $6$ can be rewritten as $2y' = 6$, while $2z' = 4$. This is the definition of even numbers.
I then get $y' = 3, z' = 2$. Now I substitute those values into the original equation. $x = 4(3)$ and $x = 6(2)$. both resulting in $12$. Have I shown in a valid way that the set $T$ contains factors of $12$, while set $S$ does not?

REPLY [1 votes]: If you have to show two sets are not equal, it suffices to find an element in one set and not the other. In this case, $12\not\in S$ but $12\in T$.
It turns out that $S\subsetneq T$, but this is not needed for the question at hand.


TITLE: Harmonic number inequality
QUESTION [11 upvotes]: With the $n$th harmonic number defined as $$
H_n  = \sum\limits_{k = 1}^n  {\frac{1}{k}}
$$ I'm supposed to find the minimum EXACT value of $n$ such that $H_n>100$ . I could only find approximations, but my professor demands an exact value . Is this possible ?
If so, how can we find it?

REPLY [9 votes]: We have:
$$ \log(n+1/2)+\frac{1}{24n^2}-\frac{1}{24n^3}\leq H_n-\gamma \leq \log(n+1/2)+\frac{1}{24n^2}$$
hence the smallest natural number for which $H_n>100$ is about $m=\left\lfloor e^{100-\gamma}-\frac{1}{2}\right\rfloor.$

Notice that $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}=-\gamma+H_{x-1}$ is a concave function, since:
$$ \psi'(x) = \sum_{n\geq 0}\frac{1}{(n+x)^2}.$$
This gives that the Newton's method with starting point $m$ is a really wise choice to find the exact value of the least integer $n$ such that $H_n>100$.

REPLY [4 votes]: The simplest pertinent estimate is based on Riemann sums. Specifically, if the step size is $h=1$, then $H_n$ is the corresponding upper Riemann sum for
$$\int_1^n \frac{1}{x} dx.$$
Hence $H_n \geq \ln(n)-\ln(1) = \ln(n)$. Solving $\ln(n) \geq 100$ we have $n \geq e^{100}$.
For the other bound, we have that
$$H_n - 1 \leq \int_1^n \frac{1}{x} dx \Rightarrow H_n \leq \ln(n)+1.$$
Solving $\ln(n)+1 \leq 100$ we have $n \leq e^{99}$, so your minimal answer will be between $e^{99}$ and $e^{100}$.
This is clearly quite crude. However, about the only way to improve this is to accurately estimate the Euler-Mascheroni constant $\gamma$, since when $n$ is large we have $H_n \approx \ln(n) + \gamma$, meaning that your minimal answer should be in the vicinity of $e^{100-\gamma}$.

REPLY [3 votes]: The answer is $$\lceil\chi\rceil$$
where $\chi$ is the solution of $$\text{N}(\chi)=100$$
where $\text{N}$ is the analytic continuation of the harmonic sum function, and $\lceil\rceil$ is the integer ceiling function.
This can be quickly calculated in Mathematica:
Ceiling[x] /. Solve[HarmonicNumber[x] == 100, x][[1]]

which instantly yields the answer:
15092688622113788323693563264538101449859497


TITLE: Find the area enclosed by the loop $r=2(1-\sin\theta)\sqrt{\cos\theta}$
QUESTION [4 upvotes]: The diagram shows a sketch of the loop whose polar equation is
$$r=2(1-\sin\theta),\qquad -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$$
a)Show that the area enclosed by the loop is 16/3.
b)Show that the initial line divides the area enclosed by the loop in the ratio 1:7.

REPLY [1 votes]: If you want the details here it is. The area of the loop has two parts$ I_1 $ and $I_2$ where. $$\int_{-\pi/2}^{0}\frac{1}{2}r^2d\theta \int_{0}^{+\pi/2}\frac{1}{2}r^2d\theta= I_1+I_2 $$ Now $$I1=\int_{-\pi/2}^{0}\frac{1}{2}r^2d\theta=\int_{-\pi/2}^{0}2(1-\sin\theta)^2\cos\theta d\theta$$. Assuming $x=\sin\theta$ and observing $dx=\cos\theta d\theta$ we have $$ I_1=\int_{-1}^{0}2(1-x)^2 dx=14/3$$ Similarly $$I_2=\int_{0}^{+1}2(1-x)^2 dx=2/3$$ Showing that $I_1+I_2=16/3$ and $I2/I1=1:7$


TITLE: Partial Derivative Disambiguation
QUESTION [4 upvotes]: There are at least two substantially different meanings to $\frac{\partial}{\partial x}f(x,\ y,\ z(x))$.  The $\partial x$ could mean "with respect to $x$ the independent variable," or it could mean "with respect to the $x$ the first parameter of $f$."  I think this can be understood in light of a net income calculation. Suppose $x$ is an individual's taxable gross income, $y$ is her non-taxable gross income (gifts received, etc.), $z$ is her income tax, and $f$ is her net income, all over the same year.  Since net income depends on taxable gross income, non-taxable gross income, and income tax, as given by $f = x + y - z$, and income tax depends on taxable gross income, as given by $z = .15x$ (using a single 15% tax bracket for simplicity), we can write the overall equation as $f(x,\ y,\ z(x)) = x + y - z(x)$ where $z(x) = .15x$, and then consider the meaning of $\frac{\partial}{\partial x}f(x,\ y,\ z(x))$.
If we interpret $\partial x$ to mean "with respect to $x$ the independent variable," then $\frac{\partial}{\partial x}f(x,\ y,\ z(x))$ represents the change in net income relative to a reported change in taxable gross income, whereas if we interpret $\partial x$ to mean "with respect to the $x$ the first parameter of $f$," then $\frac{\partial}{\partial x}f(x,\ y,\ z(x))$ represents the change in net income relative to an unreported change in taxable gross income.
I have given, just as I have learned, a binary explanation of this difference.  The $\partial x$ refers to either an independent variable or a parameter of $f$.  My question is whether it is also acceptable for it to refer to something in between.  Let's assign a new color to the contents of each nested layer of a function's parentheses, so that the above example becomes $f(\color{blue}{\textrm{x, y, z(}}\color{tan}{\textrm{x}}\color{blue}{\textrm{)}})$.  This disambiguates things by allowing us to refer to the change in net income relative to a reported change in taxable gross income with $\frac{\partial}{\partial \color{tan}{\textrm{x}}}f(\color{blue}{\textrm{x, y, z(}}\color{tan}{\textrm{x}}\color{blue}{\textrm{)}})$ and the change in net income relative to an unreported change in taxable gross income with $\frac{\partial}{\partial \color{blue}{\textrm{x}}}f(\color{blue}{\textrm{x, y, z(}}\color{tan}{\textrm{x}}\color{blue}{\textrm{)}})$.  I think colors are less misleading than subscripts in this case, because $\color{tan}{\textrm{x}}$ and $\color{blue}{\textrm{x}}$ are the same algebraic entity; it's just that when the calculus eats an algebraic expression and spits out a new one, it sometimes chews up the two $x$'s a bit differently.
With this setup, the question can be asked quite succinctly; can $\frac{\partial}{\partial x}f(x,\ y,\ z(x),\ a(x,\ z(x)), b(x,\ z(x))$ also mean $\frac{\partial}{\partial \color{orange}{\textrm{x}}}f(\color{blue}{\textrm{x, y, z(}}\color{tan}{\textrm{x}}\color{blue}{\textrm{), a(}}\color{orange}{\textrm{x, z(}}\color{tan}{\textrm{x}}\color{orange}{\textrm{)}}\color{blue}{\textrm{), b(}}\color{orange}{\textrm{x, z(}}\color{tan}{\textrm{x}}\color{orange}{\textrm{)}}\color{blue}{\textrm{)}}$ and/or $\frac{\partial}{\partial \color{lime}{\textrm{x}}}f(\color{blue}{\textrm{x, y, z(}}\color{tan}{\textrm{x}}\color{blue}{\textrm{), a(}}\color{lime}{\textrm{x, z(}}\color{tan}{\textrm{x}}\color{lime}{\textrm{)}}\color{blue}{\textrm{), b(}}\color{red}{\textrm{x, z(}}\color{tan}{\textrm{x}}\color{red}{\textrm{)}}\color{blue}{\textrm{)}}$, and/or have some other meaning drawn via a similar color hierarchy, or is a partial derivative unable to be taken with respect to orange or green (or red) $x$, since they are neither independent variables nor parameters of $f$?

REPLY [0 votes]: If you paremetrize z in terms of x, $z(x)$, then $f(x,y,z(x))$ is actually a function depending on x & y, $f(x,y)$. taking a $\partial _xf(x,y)=\partial _xf(x,y,z(x))$ if you define $\partial _xf(x,y,z(x))$ as "taking derivative with respect to (wrt) x and not z(x)" + $\partial _zf(x,y,z(x))z'(x)$. But after all, I think it is a convention, since defining $\partial_x$ as "taking derivative wrt x and not z(x)" can be useful.


TITLE: Defining real numbers to exclude incomputable numbers
QUESTION [1 upvotes]: The real numbers are normally constructed via Dedekind cuts or similar approaches, which result in incomputable numbers: numbers that no finite algorithm can produce to arbitrary precision.  This is in contrast with the computable reals, such as $\pi$, which, while irrational, can be computed to arbitrary precision via a finite algorithm (e.g. a Turing machine).
Would an alternate approach, that required a set, sequence, or number, whether finite or infinite, to have a finite definition, work?
It would seem to me that Dedekind cuts would still apply (as the sets of rationals need to be definable, e.g. "the set of rationals that, when squared, are less than the limit of $\sum \frac{x^n}{n!}$"), Completeness Property would still be met (any set with an upper bound has a least upper bound; as all sets need a finite definition, there is no set with an uncomputable bound).
Is there any inconsistency with defining sets, sequences, and the real numbers this way?  Would it violate any of the principles of real analysis?
And, if not, what reason do we have to include sets which lack finite definitions, and result in incomputable numbers?

REPLY [7 votes]: If we interpret "has a finite description" as "is computable," then in fact there are very stark differences between classical analysis and its computable analogue. For example (contra your claim), there are computable monotonically increasing bounded sequences of rationals whose suprema are not computable; such sequences are called Specker sequences. The general study of analysis from a computable perspective is, appropriately enough, called computable analysis and there are a number of textbooks and papers on the topic.
One possible response to this is to broaden the notion of "finitely describable," but at this point we need to be very careful about our intuitions around this notion. For example, we could restrict attention to the real numbers which are parameter-freely first-order definable over some large set (e.g. some "big enough" level of the cumulative hierarchy); this would guarantee lots of nice analytic properties, but to me far overshoots the notion of "finitely describable" that we're going for here.


TITLE: Combinatorics-Find number of ways we can select cells from a nXn grid such that the number of cells selected from each row and column is odd
QUESTION [2 upvotes]: Find number of ways we can select cells from a nXn grid such that the number of cells selected from each row and column is odd.
Any hints?

REPLY [4 votes]: Fill in the first $n-1$ rows (almost) arbitrarily with $0$'s and/or $1$'s, the only condition being that each row sum is odd. There are $2^{n-1}$ ways to do it for each row, for a total of $(2^{n-1})^{n-1}=2^{(n-1)^2}$.
Now fill in the last row to make all column sums odd.  There is a unique way to do this.
Finally, argue that the last row sum is odd. There are two cases, $n$ odd and $n$ even. If $n$ is odd, the total number of $1$'s is odd, because each column sum is odd.  Since the sum of the first $n-1$ rows is even, the last row sum is  odd.
If $n$ is even, then the total is even, but the sum of the first $n-1$ rows is odd, so again the the last row sum is odd.


TITLE: How many ways can 2m objects be paired and assigned to m boxes?
QUESTION [6 upvotes]: In how many ways can $2m$ objects be paired and assigned to $m$ boxes?
In that post the questioner claims that there are $\frac{(2m)!}{2^m}$ to pair $2m$ and assign them to $m$ boxes.
My attempt :-
First we need to find number of ways to pair $2m$ objects.
That is $\frac{(2m)!}{2! * (2m-2)!}$ = $\frac{2m * (2m - 1)}{2}$ = $m * (2m - 1)$
After pairing there are $m$ objects left, and there are $m!$ to permute them in $m$ boxes
Thus the answer should be $m! * m * (2m -1)$ ways.
Which is wrong. what i am doing incorrectly, please anyone correct me.

REPLY [2 votes]: Line up the $m$ boxes in some order from left to right.  We choose two of the $2m$ objects to place in the first box, choose two of the remaining $2m - 2$ boxes in the second box, choose two of the remaining $2m - 4$ boxes in the third box, and so forth until we place the final two objects in the $m$th box.  We can do this in
\begin{align*}
\binom{2m}{2}&\binom{2m - 2}{2}\binom{2m - 4}{2} \cdots \binom{2}{2}\\
& = \frac{(2m)!}{2!(2m - 2)!} \cdot \frac{(2m - 2)!}{2!(2m - 4)!} \cdot \frac{(2m - 4)!}{2!(2m - 6)!} \cdots \frac{2!}{2!0!}\\
& = \frac{(2m)!}{2^m}
\end{align*}
ways.
The number $\binom{2m}{2}$ is the number of ways to choose a particular pair of the $2m$ objects to place in a box.  It is not the number of ways to pair the objects.  To do that, we can line up the objects in a row.  There are $2m - 1$ ways we can choose an object to pair with the first object in line.  This leaves us with $2m - 2$ unpaired objects. We have $2m - 3$ ways to choose an object to pair with the first unpaired object remaining in the line.  This leaves us with $2m - 4$ unpaired objects.  We have $2m - 5$ ways to choose an object to pair with the first unpaired object remaining in the line.  Continuing in this way, we obtain
$$(2m - 1)(2m - 3)(2m - 5) \cdots 1$$
ways of placing the $2m$ objects in $m$ pairs.
Multiplying the expression for the number of pairs by $m!$ yields
\begin{align*}
(2m - 1)&(2m - 3)(2m - 5) \cdots 1 \cdot m!\\
& = (2m - 1)(2m - 3)(2m - 5) \cdots 1 \cdot m! \cdot \frac{2^m}{2^m}\\
& = (2m - 1)(2m - 3)(2m - 5) \cdots 1 \cdot \frac{(2m)(2m - 2)(2m - 4) \cdots 2}{2^m}\\
& = \frac{(2m)!}{2^m}
\end{align*}
which is the number of ways of placing the pairs in $m$ boxes.